48=-16x^2+80x

Solution for 48=-16x^2+80x equation:

48=-16x^2+80x

We move all terms to the left:

48-(-16x^2+80x)=0

We get rid of parentheses

16x^2-80x+48=0

a = 16; b = -80; c = +48;
Δ = b2-4ac
Δ = -802-4·16·48
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{13}}{2*16}=\frac{80-16\sqrt{13}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{13}}{2*16}=\frac{80+16\sqrt{13}}{32}
$