3r^2+4r-119=0

Solution for 3r^2+4r-119=0 equation:

3r^2+4r-119=0

a = 3; b = 4; c = -119;
Δ = b2-4ac
Δ = 42-4·3·(-119)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-38}{2*3}=\frac{-42}{6}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+38}{2*3}=\frac{34}{6}
=5+2/3
$