-8x^2+96x=-40

Solution for -8x^2+96x=-40 equation:

-8x^2+96x=-40

We move all terms to the left:

-8x^2+96x-(-40)=0

We add all the numbers together, and all the variables

-8x^2+96x+40=0

a = -8; b = 96; c = +40;
Δ = b2-4ac
Δ = 962-4·(-8)·40
Δ = 10496
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:

$\sqrt{\Delta}=\sqrt{10496}=\sqrt{256*41}=\sqrt{256}*\sqrt{41}=16\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{41}}{2*-8}=\frac{-96-16\sqrt{41}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{41}}{2*-8}=\frac{-96+16\sqrt{41}}{-16}
$