z2=2(3z+1)3

Solution for z2=2(3z+1)3 equation:

z2=2(3z+1)3

We move all terms to the left:

z2-(2(3z+1)3)=0

We add all the numbers together, and all the variables

z^2-(2(3z+1)3)=0


We calculate terms in parentheses: -(2(3z+1)3), so:

2(3z+1)3

We multiply parentheses

18z+6

Back to the equation:

-(18z+6)


We get rid of parentheses

z^2-18z-6=0

a = 1; b = -18; c = -6;
Δ = b2-4ac
Δ = -182-4·1·(-6)
Δ = 348
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*1}=\frac{18-2\sqrt{87}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*1}=\frac{18+2\sqrt{87}}{2}
$