b^2=13b+13=

Solution for b^2=13b+13= equation:

b^2=13b+13=

We move all terms to the left:

b^2-(13b+13)=0

We get rid of parentheses

b^2-13b-13=0

a = 1; b = -13; c = -13;
Δ = b2-4ac
Δ = -132-4·1·(-13)
Δ = 221
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{221}}{2*1}=\frac{13-\sqrt{221}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{221}}{2*1}=\frac{13+\sqrt{221}}{2}
$