z2-6=2(z+2)-10

Solution for z2-6=2(z+2)-10 equation:

z2-6=2(z+2)-10

We move all terms to the left:

z2-6-(2(z+2)-10)=0

We add all the numbers together, and all the variables

z^2-(2(z+2)-10)-6=0


We calculate terms in parentheses: -(2(z+2)-10), so:

2(z+2)-10

We multiply parentheses

2z+4-10

We add all the numbers together, and all the variables

2z-6

Back to the equation:

-(2z-6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

z^2-2z=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2}
=2
$