16x^2-72x-96=0

Solution for 16x^2-72x-96=0 equation:

16x^2-72x-96=0

a = 16; b = -72; c = -96;
Δ = b2-4ac
Δ = -722-4·16·(-96)
Δ = 11328
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:

$\sqrt{\Delta}=\sqrt{11328}=\sqrt{64*177}=\sqrt{64}*\sqrt{177}=8\sqrt{177}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{177}}{2*16}=\frac{72-8\sqrt{177}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{177}}{2*16}=\frac{72+8\sqrt{177}}{32}
$