2w^2-28w=-96

Solution for 2w^2-28w=-96 equation:

2w^2-28w=-96

We move all terms to the left:

2w^2-28w-(-96)=0

We add all the numbers together, and all the variables

2w^2-28w+96=0

a = 2; b = -28; c = +96;
Δ = b2-4ac
Δ = -282-4·2·96
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*2}=\frac{24}{4}
=6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*2}=\frac{32}{4}
=8
$