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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

b^2+82b-988=0

a = 1; b = 82; c = -988;
Δ = b2-4ac
Δ = 822-4·1·(-988)
Δ = 10676
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:

$\sqrt{\Delta}=\sqrt{10676}=\sqrt{4*2669}=\sqrt{4}*\sqrt{2669}=2\sqrt{2669}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(82)-2\sqrt{2669}}{2*1}=\frac{-82-2\sqrt{2669}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(82)+2\sqrt{2669}}{2*1}=\frac{-82+2\sqrt{2669}}{2}
$