12r^2+24r-13=0

Solution for 12r^2+24r-13=0 equation:

12r^2+24r-13=0

a = 12; b = 24; c = -13;
Δ = b2-4ac
Δ = 242-4·12·(-13)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-20\sqrt{3}}{2*12}=\frac{-24-20\sqrt{3}}{24}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+20\sqrt{3}}{2*12}=\frac{-24+20\sqrt{3}}{24}
$