First we need to factor trinomial $ \color{blue}{ -16x^2+48x+28 } $ and than we use factored form to solve an equation $ \color{blue}{ -16x^2+48x+28 = 0} $.

Step 1: We can simplify equation by multiplying both sides by -1. After multiplying we have the following equation:

$$ \begin{aligned} -16x^2+48x+28 &= 0 \,\,\, / \color{orangered}{\cdot \, -1 } \\[0.9 em ] 16x^2-48x-28 &=0 \end{aligned} $$

Step 2: Simplify equation by dividing all coefficients by 4

$$ \begin{aligned} 16x^2-48x-28 &= 0 \,\,\, / \color{orangered}{ : 4 } \\[0.9 em ] 4x^2-12x-7 &=0 \end{aligned} $$

Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 4 , b = -12 ~ \text{ and } ~ c = -7 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -7} $.

$$ a \cdot c = -28 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -28 $ and add to $ b = -12 $.

Step 4: All pairs of numbers with a product of $ -28 $ are:

PRODUCT = -28
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -12 }$

PRODUCT = -28 and SUM = -12
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 6: Replace middle term $ -12 x $ with $ 2x-14x $:

$$ 4x^{2}-12x-7 = 4x^{2}+2x-14x-7 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 4x^{2}+2x-14x-7 = 2x\left(2x+1\right) -7\left(2x+1\right) = \left(2x-7\right) \left(2x+1\right) $$

Step 8: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} 2x-7 &= 0 \\ x &= \frac{ 7 }{ 2 } \end{aligned} & ~ & \begin{aligned} 2x+1 &= 0 \\ x &= -\frac{ 1 }{ 2 } \end{aligned} \end{array} $$

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