(x^2)+8x+16=4(x+28)

Solution for (x^2)+8x+16=4(x+28) equation:

(x^2)+8x+16=4(x+28)

We move all terms to the left:

(x^2)+8x+16-(4(x+28))=0


We calculate terms in parentheses: -(4(x+28)), so:

4(x+28)

We multiply parentheses

4x+112

Back to the equation:

-(4x+112)


We get rid of parentheses

x^2+8x-4x-112+16=0

We add all the numbers together, and all the variables

x^2+4x-96=0

a = 1; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*1}=\frac{16}{2}
=8
$