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

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

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

We add all the numbers together, and all the variables

r^2+2r-3=0

a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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