4x^2+24x+36=16

Solution for 4x^2+24x+36=16 equation:

4x^2+24x+36=16

We move all terms to the left:

4x^2+24x+36-(16)=0

We add all the numbers together, and all the variables

4x^2+24x+20=0

a = 4; b = 24; c = +20;
Δ = b2-4ac
Δ = 242-4·4·20
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*4}=\frac{-40}{8}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*4}=\frac{-8}{8}
=-1
$