Precalculus Examples

$x^{2} - 7 x - 10 = - 2 x - 4$

Step 1

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

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Step 1.1

Add $2 x$ to both sides of the equation.

$x^{2} - 7 x - 10 + 2 x = - 4$

Step 1.2

Add $- 7 x$ and $2 x$ .

$x^{2} - 5 x - 10 = - 4$

Step 2

Add $4$ to both sides of the equation.

$x^{2} - 5 x - 10 + 4 = 0$

Step 3

Add $- 10$ and $4$ .

$x^{2} - 5 x - 6 = 0$

Step 4

Factor $x^{2} - 5 x - 6$ using the AC method.

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Step 4.1

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 $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Step 4.2

Write the factored form using these integers.

$(x - 6) (x + 1) = 0$

Step 5

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

$x - 6 = 0$

$x + 1 = 0$

Step 6

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

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Step 6.1

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

$x - 6 = 0$

Step 6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 7

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

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Step 7.1

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

$x + 1 = 0$

Step 7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

The final solution is all the values that make $(x - 6) (x + 1) = 0$ true.

$x = 6, - 1$