r^2+4r=5

Solution for r^2+4r=5 equation:

r^2+4r=5

We move all terms to the left:

r^2+4r-(5)=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$