-16x^2+96x=40

Solution for -16x^2+96x=40 equation:

-16x^2+96x=40

We move all terms to the left:

-16x^2+96x-(40)=0

a = -16; b = 96; c = -40;
Δ = b2-4ac
Δ = 962-4·(-16)·(-40)
Δ = 6656
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:

$\sqrt{\Delta}=\sqrt{6656}=\sqrt{256*26}=\sqrt{256}*\sqrt{26}=16\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{26}}{2*-16}=\frac{-96-16\sqrt{26}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{26}}{2*-16}=\frac{-96+16\sqrt{26}}{-32}
$