-3z+1=(2z+4)(3z-6)

Solution for -3z+1=(2z+4)(3z-6) equation:

-3z+1=(2z+4)(3z-6)

We move all terms to the left:

-3z+1-((2z+4)(3z-6))=0

We multiply parentheses ..

-((+6z^2-12z+12z-24))-3z+1=0


We calculate terms in parentheses: -((+6z^2-12z+12z-24)), so:

(+6z^2-12z+12z-24)

We get rid of parentheses

6z^2-12z+12z-24

We add all the numbers together, and all the variables

6z^2-24

Back to the equation:

-(6z^2-24)


We add all the numbers together, and all the variables

-3z-(6z^2-24)+1=0

We get rid of parentheses

-6z^2-3z+24+1=0

We add all the numbers together, and all the variables

-6z^2-3z+25=0

a = -6; b = -3; c = +25;
Δ = b2-4ac
Δ = -32-4·(-6)·25
Δ = 609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{609}}{2*-6}=\frac{3-\sqrt{609}}{-12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{609}}{2*-6}=\frac{3+\sqrt{609}}{-12}
$