64+96x-16x^2=0

Solution for 64+96x-16x^2=0 equation:

64+96x-16x^2=0

a = -16; b = 96; c = +64;
Δ = b2-4ac
Δ = 962-4·(-16)·64
Δ = 13312
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:

$\sqrt{\Delta}=\sqrt{13312}=\sqrt{1024*13}=\sqrt{1024}*\sqrt{13}=32\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{13}}{2*-16}=\frac{-96-32\sqrt{13}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{13}}{2*-16}=\frac{-96+32\sqrt{13}}{-32}
$