b^2=-35+12b

Solution for b^2=-35+12b equation:

b^2=-35+12b

We move all terms to the left:

b^2-(-35+12b)=0

We add all the numbers together, and all the variables

b^2-(12b-35)=0

We get rid of parentheses

b^2-12b+35=0

a = 1; b = -12; c = +35;
Δ = b2-4ac
Δ = -122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2}{2*1}=\frac{10}{2}
=5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2}{2*1}=\frac{14}{2}
=7
$