8z-22=3(3z+11)6z

Solution for 8z-22=3(3z+11)6z equation:

8z-22=3(3z+11)6z

We move all terms to the left:

8z-22-(3(3z+11)6z)=0


We calculate terms in parentheses: -(3(3z+11)6z), so:

3(3z+11)6z

We multiply parentheses

54z^2+198z

Back to the equation:

-(54z^2+198z)


We get rid of parentheses

-54z^2+8z-198z-22=0

We add all the numbers together, and all the variables

-54z^2-190z-22=0

a = -54; b = -190; c = -22;
Δ = b2-4ac
Δ = -1902-4·(-54)·(-22)
Δ = 31348
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:

$\sqrt{\Delta}=\sqrt{31348}=\sqrt{4*7837}=\sqrt{4}*\sqrt{7837}=2\sqrt{7837}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-190)-2\sqrt{7837}}{2*-54}=\frac{190-2\sqrt{7837}}{-108}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-190)+2\sqrt{7837}}{2*-54}=\frac{190+2\sqrt{7837}}{-108}
$