3x^2=2x^2+23

Solution for 3x^2=2x^2+23 equation:

3x^2=2x^2+23

We move all terms to the left:

3x^2-(2x^2+23)=0

We get rid of parentheses

3x^2-2x^2-23=0

We add all the numbers together, and all the variables

x^2-23=0

a = 1; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·1·(-23)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{23}}{2*1}=\frac{0-2\sqrt{23}}{2}
=-\frac{2\sqrt{23}}{2}
=-\sqrt{23}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{23}}{2*1}=\frac{0+2\sqrt{23}}{2}
=\frac{2\sqrt{23}}{2}
=\sqrt{23}
$