(z-2)(z-2)=-7(z-2)

Solution for (z-2)(z-2)=-7(z-2) equation:

(z-2)(z-2)=-7(z-2)

We move all terms to the left:

(z-2)(z-2)-(-7(z-2))=0

We multiply parentheses ..

(+z^2-2z-2z+4)-(-7(z-2))=0


We calculate terms in parentheses: -(-7(z-2)), so:

-7(z-2)

We multiply parentheses

-7z+14

Back to the equation:

-(-7z+14)


We get rid of parentheses

z^2-2z-2z+7z+4-14=0

We add all the numbers together, and all the variables

z^2+3z-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$