4r^2+17r-15=0

Solution for 4r^2+17r-15=0 equation:

4r^2+17r-15=0

a = 4; b = 17; c = -15;
Δ = b2-4ac
Δ = 172-4·4·(-15)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*4}=\frac{-40}{8}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*4}=\frac{6}{8}
=3/4
$