5b^2+17b-40=0

Solution for 5b^2+17b-40=0 equation:

5b^2+17b-40=0

a = 5; b = 17; c = -40;
Δ = b2-4ac
Δ = 172-4·5·(-40)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-33}{2*5}=\frac{-50}{10}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+33}{2*5}=\frac{16}{10}
=1+3/5
$