10(x2-49)/3x(x2-4)(x+1)=0

Solution for 10(x2-49)/3x(x2-4)(x+1)=0 equation:

10(x2-49)/3x(x2-4)(x+1)=0


Domain of the equation: 3x(x2-4)(x+1)!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


10(+x^2-49)=0

We multiply 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
$