2x^2+28x=96

Solution for 2x^2+28x=96 equation:

2x^2+28x=96

We move all terms to the left:

2x^2+28x-(96)=0

a = 2; b = 28; c = -96;
Δ = b2-4ac
Δ = 282-4·2·(-96)
Δ = 1552
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{97}}{2*2}=\frac{-28-4\sqrt{97}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{97}}{2*2}=\frac{-28+4\sqrt{97}}{4}
$