3a-3-24a-3a-4-20a+5+4a^2=0

Solution for 3a-3-24a-3a-4-20a+5+4a^2=0 equation:

3a-3-24a-3a-4-20a+5+4a^2=0

We add all the numbers together, and all the variables

4a^2-44a-2=0

a = 4; b = -44; c = -2;
Δ = b2-4ac
Δ = -442-4·4·(-2)
Δ = 1968
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:

$\sqrt{\Delta}=\sqrt{1968}=\sqrt{16*123}=\sqrt{16}*\sqrt{123}=4\sqrt{123}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{123}}{2*4}=\frac{44-4\sqrt{123}}{8}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{123}}{2*4}=\frac{44+4\sqrt{123}}{8}
$