r^2=13+4r

Solution for r^2=13+4r equation:

r^2=13+4r

We move all terms to the left:

r^2-(13+4r)=0

We add all the numbers together, and all the variables

r^2-(4r+13)=0

We get rid of parentheses

r^2-4r-13=0

a = 1; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·1·(-13)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{17}}{2*1}=\frac{4-2\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{17}}{2*1}=\frac{4+2\sqrt{17}}{2}
$