3x^2+8=9x^2+3x^2

Solution for 3x^2+8=9x^2+3x^2 equation:

3x^2+8=9x^2+3x^2

We move all terms to the left:

3x^2+8-(9x^2+3x^2)=0

We get rid of parentheses

3x^2-9x^2-3x^2+8=0

We add all the numbers together, and all the variables

-9x^2+8=0

a = -9; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-9)·8
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-9}=\frac{0-12\sqrt{2}}{-18}
=-\frac{12\sqrt{2}}{-18}
=-\frac{2\sqrt{2}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-9}=\frac{0+12\sqrt{2}}{-18}
=\frac{12\sqrt{2}}{-18}
=\frac{2\sqrt{2}}{-3}
$