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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3z^2-6z+z-2)-((z-1)(z+7))=0


We calculate terms in parentheses: -((z-1)(z+7)), so:

(z-1)(z+7)

We multiply parentheses ..

(+z^2+7z-1z-7)

We get rid of parentheses

z^2+7z-1z-7

We add all the numbers together, and all the variables

z^2+6z-7

Back to the equation:

-(z^2+6z-7)


We get rid of parentheses

3z^2-z^2-6z+z-6z-2+7=0

We add all the numbers together, and all the variables

2z^2-11z+5=0

a = 2; b = -11; c = +5;
Δ = b2-4ac
Δ = -112-4·2·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*2}=\frac{2}{4}
=1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*2}=\frac{20}{4}
=5
$