5r^2+12r-17=0

Solution for 5r^2+12r-17=0 equation:

5r^2+12r-17=0

a = 5; b = 12; c = -17;
Δ = b2-4ac
Δ = 122-4·5·(-17)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-22}{2*5}=\frac{-34}{10}
=-3+2/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+22}{2*5}=\frac{10}{10}
=1
$