(b+2)(b-3)=10

Solution for (b+2)(b-3)=10 equation:

(b+2)(b-3)=10

We move all terms to the left:

(b+2)(b-3)-(10)=0

We multiply parentheses ..

(+b^2-3b+2b-6)-10=0

We get rid of parentheses

b^2-3b+2b-6-10=0

We add all the numbers together, and all the variables

b^2-1b-16=0

a = 1; b = -1; c = -16;
Δ = b2-4ac
Δ = -12-4·1·(-16)
Δ = 65
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{65}}{2*1}=\frac{1-\sqrt{65}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*1}=\frac{1+\sqrt{65}}{2}
$