(-9/(2x^2))+9=0

Solution for (-9/(2x^2))+9=0 equation:

(-9/(2x^2))+9=0


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

x!=0/1

x!=0

x∈R


We get rid of parentheses

-9/2x^2+9=0

We multiply all the terms by the denominator


9*2x^2-9=0

Wy multiply elements

18x^2-9=0

a = 18; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·18·(-9)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*18}=\frac{0-18\sqrt{2}}{36}
=-\frac{18\sqrt{2}}{36}
=-\frac{\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*18}=\frac{0+18\sqrt{2}}{36}
=\frac{18\sqrt{2}}{36}
=\frac{\sqrt{2}}{2}
$