12r^2+23r=24

Solution for 12r^2+23r=24 equation:

12r^2+23r=24

We move all terms to the left:

12r^2+23r-(24)=0

a = 12; b = 23; c = -24;
Δ = b2-4ac
Δ = 232-4·12·(-24)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-41}{2*12}=\frac{-64}{24}
=-2+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+41}{2*12}=\frac{18}{24}
=3/4
$