80x-16x^2-96=0

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

80x-16x^2-96=0

a = -16; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·(-16)·(-96)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16}{2*-16}=\frac{-96}{-32}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16}{2*-16}=\frac{-64}{-32}
=+2
$