60r^2+4r-160=0

Solution for 60r^2+4r-160=0 equation:

60r^2+4r-160=0

a = 60; b = 4; c = -160;
Δ = b2-4ac
Δ = 42-4·60·(-160)
Δ = 38416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{38416}=196$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-196}{2*60}=\frac{-200}{120}
=-1+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+196}{2*60}=\frac{192}{120}
=1+3/5
$