(10(x^2-49))/(3x(x^2-4)(x+1))=0

Solution for (10(x^2-49))/(3x(x^2-4)(x+1))=0 equation:

(10(x^2-49))/(3x(x^2-4)(x+1))=0


Domain of the equation: (3x(x^2-4)(x+1))!=0

x∈R


We multiply all the terms by the denominator


(10(x^2-49))=0


We calculate terms in parentheses: +(10(x^2-49)), so:

10(x^2-49)

We multiply parentheses

10x^2-490

Back to the equation:

+(10x^2-490)


We get rid of parentheses

10x^2-490=0

a = 10; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·10·(-490)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*10}=\frac{-140}{20}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*10}=\frac{140}{20}
=7
$