Miriam Blair

2023-03-14

How to solve $2{x}^{2}-2=6$?

Aaliyah Padilla

Beginner2023-03-15Added 10 answers

First, to isolate the $x}^{2$ term and maintain the equation's balance, add $2$ to each side of the equation:

$2{x}^{2}-2+{2}=6+{2}$

$2{x}^{2}-0=8$

$2{x}^{2}=8$

Next, divide each side of the equation by $2$ to isolate $x}^{2$ while keeping the equation balanced:

$\frac{2{x}^{2}}{{2}}=\frac{8}{{2}}$

$\frac{{\overline{){2}}}{x}^{2}}{\overline{){2}}}=4$

${x}^{2}=4$

To solve for $x$ while keeping the equation balanced, take the square root of each side of the equation. Remember that there are two solutions—one negative and one positive—when computing the square root of a number:

$\sqrt{{x}^{2}}=\pm \sqrt{4}$

$x=\pm \sqrt{4}=\pm 2$

$2{x}^{2}-2+{2}=6+{2}$

$2{x}^{2}-0=8$

$2{x}^{2}=8$

Next, divide each side of the equation by $2$ to isolate $x}^{2$ while keeping the equation balanced:

$\frac{2{x}^{2}}{{2}}=\frac{8}{{2}}$

$\frac{{\overline{){2}}}{x}^{2}}{\overline{){2}}}=4$

${x}^{2}=4$

To solve for $x$ while keeping the equation balanced, take the square root of each side of the equation. Remember that there are two solutions—one negative and one positive—when computing the square root of a number:

$\sqrt{{x}^{2}}=\pm \sqrt{4}$

$x=\pm \sqrt{4}=\pm 2$