15r^2+34r-16=0

Solution for 15r^2+34r-16=0 equation:

15r^2+34r-16=0

a = 15; b = 34; c = -16;
Δ = b2-4ac
Δ = 342-4·15·(-16)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-46}{2*15}=\frac{-80}{30}
=-2+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*15}=\frac{12}{30}
=2/5
$