r^2+r=15

Solution for r^2+r=15 equation:

r^2+r=15

We move all terms to the left:

r^2+r-(15)=0

a = 1; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·1·(-15)
Δ = 61
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{-(1)-\sqrt{61}}{2*1}=\frac{-1-\sqrt{61}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{61}}{2*1}=\frac{-1+\sqrt{61}}{2}
$