Solve by Factoring 2x^2-50=x^2-1

$2 x^{2} - 50 = x^{2} - 1$

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$2 x^{2} - 50 - x^{2} = - 1$

Add $1$ to both sides of the equation.

$2 x^{2} - 50 - x^{2} + 1 = 0$

Simplify $2 x^{2} - 50 - x^{2} + 1$ .

Subtract $x^{2}$ from $2 x^{2}$ .

$x^{2} - 50 + 1 = 0$

Add $- 50$ and $1$ .

$x^{2} - 49 = 0$

Rewrite $49$ as $7^{2}$ .

$x^{2} - 7^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 7$ .

$(x + 7) (x - 7) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 7 = 0$

$x - 7 = 0$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

The final solution is all the values that make $(x + 7) (x - 7) = 0$ true.

$x = - 7, 7$