3r^2+4r-18=0

Solution for 3r^2+4r-18=0 equation:

3r^2+4r-18=0

a = 3; b = 4; c = -18;
Δ = b2-4ac
Δ = 42-4·3·(-18)
Δ = 232
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{58}}{2*3}=\frac{-4-2\sqrt{58}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{58}}{2*3}=\frac{-4+2\sqrt{58}}{6}
$