-3x^2+48x-180-3=0

Solution for -3x^2+48x-180-3=0 equation:

-3x^2+48x-180-3=0

We add all the numbers together, and all the variables

-3x^2+48x-183=0

a = -3; b = 48; c = -183;
Δ = b2-4ac
Δ = 482-4·(-3)·(-183)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-6\sqrt{3}}{2*-3}=\frac{-48-6\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+6\sqrt{3}}{2*-3}=\frac{-48+6\sqrt{3}}{-6}
$