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

Solution for -16x^2+80x+48=0 equation:

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

a = -16; b = 80; c = +48;
Δ = b2-4ac
Δ = 802-4·(-16)·48
Δ = 9472
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:

$\sqrt{\Delta}=\sqrt{9472}=\sqrt{256*37}=\sqrt{256}*\sqrt{37}=16\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16\sqrt{37}}{2*-16}=\frac{-80-16\sqrt{37}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16\sqrt{37}}{2*-16}=\frac{-80+16\sqrt{37}}{-32}
$