-8x^2+48x-32=0

Solution for -8x^2+48x-32=0 equation:

-8x^2+48x-32=0

a = -8; b = 48; c = -32;
Δ = b2-4ac
Δ = 482-4·(-8)·(-32)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{5}}{2*-8}=\frac{-48-16\sqrt{5}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{5}}{2*-8}=\frac{-48+16\sqrt{5}}{-16}
$