(7b^2-3b+2)+(-8b^2+6+2b)=0

Solution for (7b^2-3b+2)+(-8b^2+6+2b)=0 equation:

(7b^2-3b+2)+(-8b^2+6+2b)=0

We get rid of parentheses

-8b^2+7b^2+2b-3b+6+2=0

We add all the numbers together, and all the variables

-1b^2-1b+8=0

a = -1; b = -1; c = +8;
Δ = b2-4ac
Δ = -12-4·(-1)·8
Δ = 33
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{-(-1)-\sqrt{33}}{2*-1}=\frac{1-\sqrt{33}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{33}}{2*-1}=\frac{1+\sqrt{33}}{-2}
$