15r^2-22r-5=0

Solution for 15r^2-22r-5=0 equation:

15r^2-22r-5=0

a = 15; b = -22; c = -5;
Δ = b2-4ac
Δ = -222-4·15·(-5)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-28}{2*15}=\frac{-6}{30}
=-1/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+28}{2*15}=\frac{50}{30}
=1+2/3
$