(6-z)(3z+1)=1

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

(6-z)(3z+1)=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3z^2-1z+18z+6)-1=0

We get rid of parentheses

-3z^2-1z+18z+6-1=0

We add all the numbers together, and all the variables

-3z^2+17z+5=0

a = -3; b = 17; c = +5;
Δ = b2-4ac
Δ = 172-4·(-3)·5
Δ = 349
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{349}}{2*-3}=\frac{-17-\sqrt{349}}{-6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{349}}{2*-3}=\frac{-17+\sqrt{349}}{-6}
$