(6z-9z^2)/(3z)+2(3-4z)=19

Solution for (6z-9z^2)/(3z)+2(3-4z)=19 equation:

(6z-9z^2)/(3z)+2(3-4z)=19

We move all terms to the left:

(6z-9z^2)/(3z)+2(3-4z)-(19)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

(6z-9z^2)/3z+2(-4z+3)-19=0

We multiply parentheses

(6z-9z^2)/3z-8z+6-19=0

We multiply all the terms by the denominator


(6z-9z^2)-8z*3z+6*3z-19*3z=0

Wy multiply elements

(6z-9z^2)-24z^2+18z-57z=0

We get rid of parentheses

-9z^2-24z^2+6z+18z-57z=0

We add all the numbers together, and all the variables

-33z^2-33z=0

a = -33; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·(-33)·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-33}{2*-33}=\frac{0}{-66}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*-33}=\frac{66}{-66}
=-1
$