-10a^2+24a+96=0

Solution for -10a^2+24a+96=0 equation:

-10a^2+24a+96=0

a = -10; b = 24; c = +96;
Δ = b2-4ac
Δ = 242-4·(-10)·96
Δ = 4416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4416}=\sqrt{64*69}=\sqrt{64}*\sqrt{69}=8\sqrt{69}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{69}}{2*-10}=\frac{-24-8\sqrt{69}}{-20}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{69}}{2*-10}=\frac{-24+8\sqrt{69}}{-20}
$