3x^2+x=16

Solution for 3x^2+x=16 equation:

3x^2+x=16

We move all terms to the left:

3x^2+x-(16)=0

a = 3; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·3·(-16)
Δ = 193
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{193}}{2*3}=\frac{-1-\sqrt{193}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*3}=\frac{-1+\sqrt{193}}{6}
$