2+z=6z2

Solution for 2+z=6z2 equation:

2+z=6z^2

We move all terms to the left:

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

determiningTheFunctionDomain
-6z^2+z+2=0

a = -6; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-6)·2
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*-6}=\frac{-8}{-12}
=2/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-6}=\frac{6}{-12}
=-1/2
$