r^2+3r=2=0

Solution for r^2+3r=2=0 equation:

r^2+3r=2=0

We move all terms to the left:

r^2+3r-(2)=0

a = 1; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·1·(-2)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{17}}{2*1}=\frac{-3-\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17}}{2*1}=\frac{-3+\sqrt{17}}{2}
$