Question

2x² = 80 - 3x²

Explanation:

Step1: Add \(3x^2\) to both sides

To solve for \(x^2\), we first eliminate the \(-3x^2\) on the right side by adding \(3x^2\) to both sides of the equation \(2x^2 = 80 - 3x^2\). This gives us \(2x^2 + 3x^2 = 80 - 3x^2 + 3x^2\). Simplifying the left side: \(2x^2+3x^2 = 5x^2\), and the right side: \(-3x^2 + 3x^2 = 0\), so the equation becomes \(5x^2=80\).

Step2: Divide both sides by 5

Now, to solve for \(x^2\), we divide both sides of the equation \(5x^2 = 80\) by 5. So, \(\frac{5x^2}{5}=\frac{80}{5}\), which simplifies to \(x^2 = 16\).

Step3: Take square root of both sides

To find \(x\), we take the square root of both sides of the equation \(x^2 = 16\). Remember that when we take the square root of a number, we get both a positive and a negative solution. So, \(x=\pm\sqrt{16}\), and \(\sqrt{16}=4\), so \(x = 4\) or \(x=-4\).

Answer:

\(x = 4\) or \(x=-4\) (or in boxed form, if we consider each solution: \(\boxed{4}\) and \(\boxed{-4}\))