z^2-2z-48=(z-8)

Solution for z^2-2z-48=(z-8) equation:

z^2-2z-48=(z-8)

We move all terms to the left:

z^2-2z-48-((z-8))=0


We calculate terms in parentheses: -((z-8)), so:

(z-8)

We get rid of parentheses

z-8

Back to the equation:

-(z-8)


We get rid of parentheses

z^2-2z-z+8-48=0

We add all the numbers together, and all the variables

z^2-3z-40=0

a = 1; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2}
=8
$