(3z+2)(z-1)=-(7z-7)

Solution for (3z+2)(z-1)=-(7z-7) equation:

(3z+2)(z-1)=-(7z-7)

We move all terms to the left:

(3z+2)(z-1)-(-(7z-7))=0

We multiply parentheses ..

(+3z^2-3z+2z-2)-(-(7z-7))=0


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

-(7z-7)

We get rid of parentheses

-7z+7

Back to the equation:

-(-7z+7)


We get rid of parentheses

3z^2-3z+2z+7z-2-7=0

We add all the numbers together, and all the variables

3z^2+6z-9=0

a = 3; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·3·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*3}=\frac{-18}{6}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*3}=\frac{6}{6}
=1
$