15r2+28r-32=0

Solution for 15r2+28r-32=0 equation:

15r^2+28r-32=0

a = 15; b = 28; c = -32;
Δ = b2-4ac
Δ = 282-4·15·(-32)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*15}=\frac{-80}{30}
=-2+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*15}=\frac{24}{30}
=4/5
$