2/3x^2-5=23

Solution for 2/3x^2-5=23 equation:

2/3x^2-5=23

We move all terms to the left:

2/3x^2-5-(23)=0


Domain of the equation: 3x^2!=0

x^2!=0/3

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x^2-28=0

We multiply all the terms by the denominator


-28*3x^2+2=0

Wy multiply elements

-84x^2+2=0

a = -84; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-84)·2
Δ = 672
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:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{42}}{2*-84}=\frac{0-4\sqrt{42}}{-168}
=-\frac{4\sqrt{42}}{-168}
=-\frac{\sqrt{42}}{-42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{42}}{2*-84}=\frac{0+4\sqrt{42}}{-168}
=\frac{4\sqrt{42}}{-168}
=\frac{\sqrt{42}}{-42}
$