15r^2=-17r-4

Solution for 15r^2=-17r-4 equation:

15r^2=-17r-4

We move all terms to the left:

15r^2-(-17r-4)=0

We get rid of parentheses

15r^2+17r+4=0

a = 15; b = 17; c = +4;
Δ = b2-4ac
Δ = 172-4·15·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*15}=\frac{-24}{30}
=-4/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*15}=\frac{-10}{30}
=-1/3
$