18a^2+19a-12=0

Solution for 18a^2+19a-12=0 equation:

18a^2+19a-12=0

a = 18; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·18·(-12)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-35}{2*18}=\frac{-54}{36}
=-1+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+35}{2*18}=\frac{16}{36}
=4/9
$