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

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

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


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

x∈R


We multiply all the terms by the denominator


(-3x^2-2x+3)=0

We get rid of parentheses

-3x^2-2x+3=0

a = -3; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-3)·3
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{10}}{2*-3}=\frac{2-2\sqrt{10}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*-3}=\frac{2+2\sqrt{10}}{-6}
$