12j^2+11j+2=0

Solution for 12j^2+11j+2=0 equation:

12j^2+11j+2=0

a = 12; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·12·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*12}=\frac{-16}{24}
=-2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*12}=\frac{-6}{24}
=-1/4
$