6z(1/2+2)+3=15

Solution for 6z(1/2+2)+3=15 equation:

6z(1/2+2)+3=15

We move all terms to the left:

6z(1/2+2)+3-(15)=0

We add all the numbers together, and all the variables

6z(1/2+2)-12=0

We multiply parentheses

6z^2+12z-12=0

a = 6; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·6·(-12)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{3}}{2*6}=\frac{-12-12\sqrt{3}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{3}}{2*6}=\frac{-12+12\sqrt{3}}{12}
$