4r^2+20=-9r+3r^2

Solution for 4r^2+20=-9r+3r^2 equation:

4r^2+20=-9r+3r^2

We move all terms to the left:

4r^2+20-(-9r+3r^2)=0

We get rid of parentheses

4r^2-3r^2+9r+20=0

We add all the numbers together, and all the variables

r^2+9r+20=0

a = 1; b = 9; c = +20;
Δ = b2-4ac
Δ = 92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*1}=\frac{-10}{2}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*1}=\frac{-8}{2}
=-4
$