96=-16x^2+80x

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

96=-16x^2+80x

We move all terms to the left:

96-(-16x^2+80x)=0

We get rid of parentheses

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