2b^2=-10+10b

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

2b^2=-10+10b

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2b^2-10b+10=0

a = 2; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·2·10
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*2}=\frac{10-2\sqrt{5}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*2}=\frac{10+2\sqrt{5}}{4}
$