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This step-by-step calculator solves quadratic equations using three different methods: the quadratic formula method, completing the square, and the factoring method. Calculator shows all the work and provides detailed explanation on how to solve an equation.

Solve $2x^2+8x-10 = 0$ by completing the square.

solution

$$\color{blue}{ x_1 = -5 }~~ \text{and}~~ \color{blue}{ x_2 = 1 }$$

explanation

Step 1: Divide equation by a number in front of the squared term. In this case we will divide by $2$.

\begin{aligned} 2x^2+8x-10 &= 0 \,\,\, ( \color{orangered}{ : 2 } ) \\[0.9 em ] x^2+4x-5 &=0 \end{aligned}

Step 2: Keep all terms containing $x$ on one side. Move $-5$ to the right.

$$x^2+4x = 5$$

Step 3: Take half of the x -term coefficient and square it. Add this value to both sides.

The x-term coefficient = $4$

The half of the x-term coefficient = $2$

After squaring we have $2^2 = 4$

When we add $4$ to both sides we have:

$$x^2+4x+4 = 5 + 4$$

Step 4: Simplify right side.

$$x^2+4x+4 = 9$$

Step 5: Write the perfect square on the left.

$$\left(x + 2 \right)^2 = 9$$

Step 6: Take the square root of both sides.

$$x + 2 = \pm \sqrt { 9 }$$

Step 7: Solve for $x$.

$x_1,x_2 = - 2 \pm \sqrt{ 9 }$

that is,

$x_1 = -5$

$x_2 = 1$

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Form values: 2 , 1 , 8 , 2 , 10 , cs , g , Solve quadratic equation 2x^2+8x-10 = 0 , , , Solve $2x^2+8x-10 = 0$ by completing the square.

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$\frac{1}{2}x^2+\frac{3}{4}x - 11 = 0$
$x^2+\sqrt{2}x - 4 = 0$
 Factoring Completing the square Quadratic formula
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EXAMPLES
example 1:ex 1:
Solve for $x^2 + 3x - 4 = 0$ by factoring.
example 2:ex 2:
Solve $4x^2 - x - 3 = 0$ by completing the square.
example 3:ex 3:
Solve $-2x^2 - 0.5x + 0.75 = 0$ using the quadratic formula.
example 4:ex 4:
Solve $\frac{2}{3} x^2 - \frac{1}{3} x - 5 = 0$.
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TUTORIAL

How to use this calculator

The most commonly used methods for solving quadratic equations are:

1. Factoring method

2. Solving quadratic equations by completing the square

In the following sections, we'll go over these methods.

Method 1A : Factoring method

If a quadratic trinomial can be factored, this is the best solving method.

We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.

Example 01: Solve $x^2 \color{red}{-8}x \color{blue}{+ 15} = 0$ by factoring.

Here we see that the leading coefficient is 1, so the factoring method is our first choice.

To factor this equation, we must find two numbers ( $a$ and $b$ ) with a sum is $a + b = \color{red}{8}$ and a product of $a \cdot b = \color{blue}{15}$.

After some trials and errors, we see that $a = 3$ and $b = 5$.

Now we use formula $x^2 - 8x + 15 = (x - a)(x - b)$ to get factored form:

$$x^2 - 8x + 15 = (x - 3)(x - 5)$$

Divide the factored form into two linear equations to get solutions.

\begin{aligned} x^2 - 8x + 15 &= 0 \\ (x - 3)(x - 5) &= 0 \\ x -3 &= 0 ~~ \text{or} ~~ x - 5= 0 \\ x &= 3 ~~ \text{or} ~~ x = 5 \end{aligned}

Method 1B : Factoring - special cases

Example 02: Solve $x^2 -8x = 0$ by factoring.

In this case, (when the coefficient c = 0 ) we can factor out $\color{blue}{x}$ out of $x^2 - 8x$.

\begin{aligned} x^2 - 8x &= 0 \\ \color{blue}{x} \cdot ( x - 8 ) &= 0 \\ x &= 0 ~~ \text{or} ~~ x - 8 = 0 \\ x &= 0 ~~ \text{or} ~~ x = 8 \end{aligned}

Example 03: Solve $x^2 - 16 = 0$ by factoring.

In this case, ( when the middle term is equal 0) we can use the difference of squares formula.

\begin{aligned} x^2 - 16 &= 0 \\ x^2 - 4^2 &= 0 \text{ use } a^2 - b^2 = (a-b)(a+b) \\ (x - 4)(x+4) &= 0 \\ x - 4 &= 0 ~~ \text{or} ~~ x + 4 = 0 \\ x &= 4 ~~ \text{or} ~~ x = -4 \end{aligned}

Method 3 : Solve using quadratic formula

This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.

Example 05: Solve equation $2x^2 + 3x - 2 = 0$ by using quadratic formula.

Step 1: Read the values of $a$, $b$, and $c$ from the quadratic equation. ( $a$ is the number in front of $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)

$$a = 2 ~~ b = 3 ~~ \text{and} ~~ c = -2$$

Step 2:Plug the values for a, b, and c into the quadratic formula and simplify.

\begin{aligned} x_1, x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-2) }}{2 \cdot 2} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{9+ 16 }}{4} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{25}}{4} \\ x_1, x_2 &= \frac{-3 \pm 5}{4} \end{aligned}

Step 3: Solve for $x_1$ and $x_2$

\begin{aligned} x_1 = & \frac{-3 \color{blue}{+} 5}{4} = \frac{2}{4} = \frac{1}{2} \\ x_2 = & \frac{-3 \color{blue}{-} 5}{4} = \frac{-8}{4} = -2 \end{aligned}

Method 2 : Completing the square

This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.

Example 04: Solve equation $2x^2 + 8x - 10= 0$ by completing the square.

Step 1: Divide the equation by the number in front of the square term.

\begin{aligned} 2x^2 + 8x - 10 & = 0 ~~ / ~ \color{orangered}{:2} \\ \frac{2x^2}{2} + \frac{8x}{2} - \frac{10}{2} & = \frac{0}{2} \\ x^2 + 4x - 5 & = 0 \end{aligned}

Step 2: move $-5$ to the right:

$$x^2 + 4x = 5$$

Step 3: Take half of the x-term coefficient $\color{blue}{\dfrac{4}{2}}$, square it $\color{blue}{\left(\dfrac{4}{2} \right)^2}$ and add this value to both sides.

$$x^2 + 4x + \color{blue}{\left(\frac{4}{2} \right)^2} = 5 + \color{blue}{\left(\frac{4}{2} \right)^2}$$

Step 4: Simplify left and right side.

$$x^2 + 4x + 2^2 = 9$$

Step 5: Write the perfect square on the left.

$$\left( x + 2 \right)^2 = 9$$

Step 6: Take the square root of both sides.

\begin{aligned} x + 2 &= \pm \sqrt{9} \\\\ x + 2 &= \pm 3 \end{aligned}

Step 7: Solve for $x_1$ and $x_2$ .

\begin{aligned} x_1 & = +3 - 2 = 1 \\ x_2 & = -3 - 2 = - 5 \end{aligned}
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