2x^2+12x+16=96

Solution for 2x^2+12x+16=96 equation:

2x^2+12x+16=96

We move all terms to the left:

2x^2+12x+16-(96)=0

We add all the numbers together, and all the variables

2x^2+12x-80=0

a = 2; b = 12; c = -80;
Δ = b2-4ac
Δ = 122-4·2·(-80)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*2}=\frac{-40}{4}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*2}=\frac{16}{4}
=4
$