First we need to factor trinomial $ \color{blue}{ 11x^2-99x+220 } $ and than we use factored form to solve an equation $ \color{blue}{ 11x^2-99x+220 = 0} $.

Step 1: Simplify equation by dividing all coefficients by 11

$$ \begin{aligned} 11x^2-99x+220 &= 0 \,\,\, / \color{orangered}{ : 11 } \\[0.9 em ] x^2-9x+20 &=0 \end{aligned} $$

Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -9 } ~ \text{ and } ~ \color{red}{ c = 20 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -9 } $ and multiply to $ \color{red}{ 20 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 20 }$.

PRODUCT = 20
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 3: Find out which pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = 20 and SUM = -9
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 4: Put -4 and -5 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-9x+20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-9x+20 & = (x -4)(x -5) \end{aligned} $$

Step 5: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} x-4 &= 0 \\ x &= 4 \end{aligned} & ~ & \begin{aligned} x-5 &= 0 \\ x &= 5 \end{aligned} \end{array} $$

Quadratic Equation Solver