(5x(x-9))/(6x^2-41x-7)=0

Solution for (5x(x-9))/(6x^2-41x-7)=0 equation:

(5x(x-9))/(6x^2-41x-7)=0


Domain of the equation: (6x^2-41x-7)!=0

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

6x^2-41x!=7

x∈R


We multiply all the terms by the denominator


(5x(x-9))=0


We calculate terms in parentheses: +(5x(x-9)), so:

5x(x-9)

We multiply parentheses

5x^2-45x

Back to the equation:

+(5x^2-45x)


We get rid of parentheses

5x^2-45x=0

a = 5; b = -45; c = 0;
Δ = b2-4ac
Δ = -452-4·5·0
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-45}{2*5}=\frac{0}{10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+45}{2*5}=\frac{90}{10}
=9
$