Solve by Factoring x^2-9x+19=x-5

$x^{2} - 9 x + 19 = x - 5$

Step 1

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

Subtract $x$ from both sides of the equation.

$x^{2} - 9 x + 19 - x = - 5$

Add $5$ to both sides of the equation.

$x^{2} - 9 x + 19 - x + 5 = 0$

Step 2

Simplify $x^{2} - 9 x + 19 - x + 5$ .

Subtract $x$ from $- 9 x$ .

$x^{2} - 10 x + 19 + 5 = 0$

Add $19$ and $5$ .

$x^{2} - 10 x + 24 = 0$

Step 3

Factor $x^{2} - 10 x + 24$ 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 $24$ and whose sum is $- 10$ .

$- 6, - 4$

Write the factored form using these integers.

$(x - 6) (x - 4) = 0$

Step 4

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

$x - 6 = 0$

$x - 4 = 0$

Step 5

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 6

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 7

The final solution is all the values that make $(x - 6) (x - 4) = 0$ true.

$x = 6, 4$

Solve by Factoring x^2-9x+19=x-5

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