6z(1/2+2)+3=13

Solution for 6z(1/2+2)+3=13 equation:

6z(1/2+2)+3=13

We move all terms to the left:

6z(1/2+2)+3-(13)=0

We add all the numbers together, and all the variables

6z(1/2+2)-10=0

We multiply parentheses

6z^2+12z-10=0

a = 6; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·6·(-10)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{6}}{2*6}=\frac{-12-8\sqrt{6}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{6}}{2*6}=\frac{-12+8\sqrt{6}}{12}
$