((3x-4)(x+7))/(9x)=0

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Solution for ((3x-4)(x+7))/(9x)=0 equation: 



((3x-4)(x+7))/(9x)=0



Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R



We multiply parentheses ..

((+3x^2+21x-4x-28))/9x=0

We multiply all the terms by the denominator


((+3x^2+21x-4x-28))=0



We calculate terms in parentheses: +((+3x^2+21x-4x-28)), so:

(+3x^2+21x-4x-28)

We get rid of parentheses

3x^2+21x-4x-28

We add all the numbers together, and all the variables

3x^2+17x-28

Back to the equation:

+(3x^2+17x-28)



We get rid of parentheses

3x^2+17x-28=0

a = 3; b = 17; c = -28;
Δ = b2-4ac
Δ = 172-4·3·(-28)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-25}{2*3}=\frac{-42}{6}
=-7
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+25}{2*3}=\frac{8}{6}
=1+1/3
$

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