3x2+x=12

Solution for 3x2+x=12 equation:

3x^2+x=12

We move all terms to the left:

3x^2+x-(12)=0

a = 3; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·3·(-12)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{145}}{2*3}=\frac{-1-\sqrt{145}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*3}=\frac{-1+\sqrt{145}}{6}
$