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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

6z^2-1z+12z+3=0

We add all the numbers together, and all the variables

6z^2+11z+3=0

a = 6; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·6·3
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{-(11)-7}{2*6}=\frac{-18}{12}
=-1+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*6}=\frac{-4}{12}
=-1/3
$