-16x2+80x=48

Solution for -16x2+80x=48 equation:

-16x^2+80x=48

We move all terms to the left:

-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}
$