6(b-5)+b2=82

Solution for 6(b-5)+b2=82 equation:

6(b-5)+b2=82

We move all terms to the left:

6(b-5)+b2-(82)=0

We add all the numbers together, and all the variables

b^2+6(b-5)-82=0

We multiply parentheses

b^2+6b-30-82=0

We add all the numbers together, and all the variables

b^2+6b-112=0

a = 1; b = 6; c = -112;
Δ = b2-4ac
Δ = 62-4·1·(-112)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-22}{2*1}=\frac{-28}{2}
=-14
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*1}=\frac{16}{2}
=8
$