7(b-5)+5b(b+4)=45

Solution for 7(b-5)+5b(b+4)=45 equation:

7(b-5)+5b(b+4)=45

We move all terms to the left:

7(b-5)+5b(b+4)-(45)=0

We multiply parentheses

5b^2+7b+20b-35-45=0

We add all the numbers together, and all the variables

5b^2+27b-80=0

a = 5; b = 27; c = -80;
Δ = b2-4ac
Δ = 272-4·5·(-80)
Δ = 2329
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{-(27)-\sqrt{2329}}{2*5}=\frac{-27-\sqrt{2329}}{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{2329}}{2*5}=\frac{-27+\sqrt{2329}}{10}
$