4x^2+24x+36-11=13

Solution for 4x^2+24x+36-11=13 equation:

4x^2+24x+36-11=13

We move all terms to the left:

4x^2+24x+36-11-(13)=0

We add all the numbers together, and all the variables

4x^2+24x+12=0

a = 4; b = 24; c = +12;
Δ = b2-4ac
Δ = 242-4·4·12
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{6}}{2*4}=\frac{-24-8\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{6}}{2*4}=\frac{-24+8\sqrt{6}}{8}
$