8z+2=z(z-5)

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

8z+2=z(z-5)

We move all terms to the left:

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


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

z(z-5)

We multiply parentheses

z^2-5z

Back to the equation:

-(z^2-5z)


We get rid of parentheses

-z^2+8z+5z+2=0

We add all the numbers together, and all the variables

-1z^2+13z+2=0

a = -1; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·(-1)·2
Δ = 177
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{-(13)-\sqrt{177}}{2*-1}=\frac{-13-\sqrt{177}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{177}}{2*-1}=\frac{-13+\sqrt{177}}{-2}
$