6z(1/2+2)+3=14

Solution for 6z(1/2+2)+3=14 equation:

6z(1/2+2)+3=14

We move all terms to the left:

6z(1/2+2)+3-(14)=0

We add all the numbers together, and all the variables

6z(1/2+2)-11=0

We multiply parentheses

6z^2+12z-11=0

a = 6; b = 12; c = -11;
Δ = b2-4ac
Δ = 122-4·6·(-11)
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{102}}{2*6}=\frac{-12-2\sqrt{102}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{102}}{2*6}=\frac{-12+2\sqrt{102}}{12}
$