2r^2+34r-120=0

Solution for 2r^2+34r-120=0 equation:

2r^2+34r-120=0

a = 2; b = 34; c = -120;
Δ = b2-4ac
Δ = 342-4·2·(-120)
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*2}=\frac{-80}{4}
=-20
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*2}=\frac{12}{4}
=3
$