12r^2-38r-12=0

Solution for 12r^2-38r-12=0 equation:

12r^2-38r-12=0

a = 12; b = -38; c = -12;
Δ = b2-4ac
Δ = -382-4·12·(-12)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{505}}{2*12}=\frac{38-2\sqrt{505}}{24}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{505}}{2*12}=\frac{38+2\sqrt{505}}{24}
$