16x^2+80x-96=0

Solution for 16x^2+80x-96=0 equation:

16x^2+80x-96=0

a = 16; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·16·(-96)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-112}{2*16}=\frac{-192}{32}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+112}{2*16}=\frac{32}{32}
=1
$