3x^2+9x+20=2x^2+2x+8

Solution for 3x^2+9x+20=2x^2+2x+8 equation:

3x^2+9x+20=2x^2+2x+8

We move all terms to the left:

3x^2+9x+20-(2x^2+2x+8)=0

We get rid of parentheses

3x^2-2x^2+9x-2x-8+20=0

We add all the numbers together, and all the variables

x^2+7x+12=0

a = 1; b = 7; c = +12;
Δ = b2-4ac
Δ = 72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*1}=\frac{-6}{2}
=-3
$