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(15x^2-24x+9)/(3x-3)=0
Domain of the equation: (3x-3)!=0We multiply all the terms by the denominator
We move all terms containing x to the left, all other terms to the right
3x!=3
x!=3/3
x!=1
x∈R
(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 $
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