b^2-10b=-10

Solution for b^2-10b=-10 equation:

b^2-10b=-10

We move all terms to the left:

b^2-10b-(-10)=0

We add all the numbers together, and all the variables

b^2-10b+10=0

a = 1; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·1·10
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{15}}{2*1}=\frac{10-2\sqrt{15}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{15}}{2*1}=\frac{10+2\sqrt{15}}{2}
$