3z^2+z=3

Solution for 3z^2+z=3 equation:

3z^2+z=3

We move all terms to the left:

3z^2+z-(3)=0

a = 3; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·3·(-3)
Δ = 37
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{37}}{2*3}=\frac{-1-\sqrt{37}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{37}}{2*3}=\frac{-1+\sqrt{37}}{6}
$