(15x2-24x+9)/(3x-3)=

Solution for (15x2-24x+9)/(3x-3)= equation:

(15x^2-24x+9)/(3x-3)=

We move all terms to the left:

(15x^2-24x+9)/(3x-3)-()=0


Domain of the equation: (3x-3)!=0

We move all terms containing x to the left, all other terms to the right

3x!=3

x!=3/3

x!=1

x∈R


We add all the numbers together, and all the variables

(15x^2-24x+9)/(3x-3)=0

We multiply all the terms by the denominator


(15x^2-24x+9)=0

We get rid of parentheses

15x^2-24x+9=0

a = 15; b = -24; c = +9;
Δ = b2-4ac
Δ = -242-4·15·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6}{2*15}=\frac{18}{30}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6}{2*15}=\frac{30}{30}
=1
$