b^2+88=2b^2-12

Solution for b^2+88=2b^2-12 equation:

b^2+88=2b^2-12

We move all terms to the left:

b^2+88-(2b^2-12)=0

We get rid of parentheses

b^2-2b^2+12+88=0

We add all the numbers together, and all the variables

-1b^2+100=0

a = -1; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-1)·100
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-1}=\frac{-20}{-2}
=+10
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-1}=\frac{20}{-2}
=-10
$