(b-7)(b+10)=60

Solution for (b-7)(b+10)=60 equation:

(b-7)(b+10)=60

We move all terms to the left:

(b-7)(b+10)-(60)=0

We multiply parentheses ..

(+b^2+10b-7b-70)-60=0

We get rid of parentheses

b^2+10b-7b-70-60=0

We add all the numbers together, and all the variables

b^2+3b-130=0

a = 1; b = 3; c = -130;
Δ = b2-4ac
Δ = 32-4·1·(-130)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*1}=\frac{-26}{2}
=-13
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*1}=\frac{20}{2}
=10
$