3x2+20x+36=4

Solution for 3x2+20x+36=4 equation:

3x^2+20x+36=4

We move all terms to the left:

3x^2+20x+36-(4)=0

We add all the numbers together, and all the variables

3x^2+20x+32=0

a = 3; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·3·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*3}=\frac{-16}{6}
=-2+2/3
$