3r^2+r=24

Solution for 3r^2+r=24 equation:

3r^2+r=24

We move all terms to the left:

3r^2+r-(24)=0

a = 3; b = 1; c = -24;
Δ = b2-4ac
Δ = 12-4·3·(-24)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*3}=\frac{-18}{6}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*3}=\frac{16}{6}
=2+2/3
$