5r^2-17r-40=0

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

5r^2-17r-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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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