Solve for x 2x^2-20x=x^2-19

$2 x^{2} - 20 x = x^{2} - 19$

Move all terms containing $x$ to the left side of the equation.

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

$2 x^{2} - 20 x - x^{2} = - 19$

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

$x^{2} - 20 x = - 19$

Add $19$ to both sides of the equation.

$x^{2} - 20 x + 19 = 0$

Factor $x^{2} - 20 x + 19$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $19$ and whose sum is $- 20$ .

$- 19, - 1$

Write the factored form using these integers.

$(x - 19) (x - 1) = 0$

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

$x - 19 = 0$

$x - 1 = 0$

Set $x - 19$ equal to $0$ and solve for $x$ .

Set $x - 19$ equal to $0$ .

$x - 19 = 0$

Add $19$ to both sides of the equation.

$x = 19$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 19) (x - 1) = 0$ true.

$x = 19, 1$

Solve for x 2x^2-20x=x^2-19

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