(b+4)(b+2)-90=0

Solution for (b+4)(b+2)-90=0 equation:

(b+4)(b+2)-90=0

We multiply parentheses ..

(+b^2+2b+4b+8)-90=0

We get rid of parentheses

b^2+2b+4b+8-90=0

We add all the numbers together, and all the variables

b^2+6b-82=0

a = 1; b = 6; c = -82;
Δ = b2-4ac
Δ = 62-4·1·(-82)
Δ = 364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{91}}{2*1}=\frac{-6-2\sqrt{91}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{91}}{2*1}=\frac{-6+2\sqrt{91}}{2}
$