-4a^2+48a-48=0

Solution for -4a^2+48a-48=0 equation:

-4a^2+48a-48=0

a = -4; b = 48; c = -48;
Δ = b2-4ac
Δ = 482-4·(-4)·(-48)
Δ = 1536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{6}}{2*-4}=\frac{-48-16\sqrt{6}}{-8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{6}}{2*-4}=\frac{-48+16\sqrt{6}}{-8}
$