12r^2+5r-2=0

Solution for 12r^2+5r-2=0 equation:

12r^2+5r-2=0

a = 12; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·12·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*12}=\frac{-16}{24}
=-2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*12}=\frac{6}{24}
=1/4
$