3r^2+34r-80=0

Solution for 3r^2+34r-80=0 equation:

3r^2+34r-80=0

a = 3; b = 34; c = -80;
Δ = b2-4ac
Δ = 342-4·3·(-80)
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*3}=\frac{-80}{6}
=-13+1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*3}=\frac{12}{6}
=2
$