r2+5r+3r2-2r=0

Solution for r2+5r+3r2-2r=0 equation:

r2+5r+3r^2-2r=0

We add all the numbers together, and all the variables

4r^2+3r=0

a = 4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*4}=\frac{-6}{8}
=-3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*4}=\frac{0}{8}
=0
$