8z+2=z(z-5)-z

Solution for 8z+2=z(z-5)-z equation:

8z+2=z(z-5)-z

We move all terms to the left:

8z+2-(z(z-5)-z)=0


We calculate terms in parentheses: -(z(z-5)-z), so:

z(z-5)-z

We add all the numbers together, and all the variables

-1z+z(z-5)

We multiply parentheses

z^2-1z-5z

We add all the numbers together, and all the variables

z^2-6z

Back to the equation:

-(z^2-6z)


We get rid of parentheses

-z^2+8z+6z+2=0

We add all the numbers together, and all the variables

-1z^2+14z+2=0

a = -1; b = 14; c = +2;
Δ = b2-4ac
Δ = 142-4·(-1)·2
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{51}}{2*-1}=\frac{-14-2\sqrt{51}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{51}}{2*-1}=\frac{-14+2\sqrt{51}}{-2}
$