2z2+z=6

Solution for 2z2+z=6 equation:

2z^2+z=6

We move all terms to the left:

2z^2+z-(6)=0

a = 2; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·2·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*2}=\frac{-8}{4}
=-2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*2}=\frac{6}{4}
=1+1/2
$