(2b-5)+b+b(b+80)=983

Solution for (2b-5)+b+b(b+80)=983 equation:

(2b-5)+b+b(b+80)=983

We move all terms to the left:

(2b-5)+b+b(b+80)-(983)=0

We add all the numbers together, and all the variables

b+(2b-5)+b(b+80)-983=0

We multiply parentheses

b^2+b+(2b-5)+80b-983=0

We get rid of parentheses

b^2+b+2b+80b-5-983=0

We add all the numbers together, and all the variables

b^2+83b-988=0

a = 1; b = 83; c = -988;
Δ = b2-4ac
Δ = 832-4·1·(-988)
Δ = 10841
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{-(83)-\sqrt{10841}}{2*1}=\frac{-83-\sqrt{10841}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(83)+\sqrt{10841}}{2*1}=\frac{-83+\sqrt{10841}}{2}
$