Solve for x 2x^2+10=4x+40

$2 x^{2} + 10 = 4 x + 40$

Subtract $4 x$ from both sides of the equation.

$2 x^{2} + 10 - 4 x = 40$

Subtract $40$ from both sides of the equation.

$2 x^{2} + 10 - 4 x - 40 = 0$

Subtract $40$ from $10$ .

$2 x^{2} - 4 x - 30 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 4 x - 30$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 4 x - 30 = 0$

Factor $2$ out of $- 4 x$ .

$2 (x^{2}) + 2 (- 2 x) - 30 = 0$

Factor $2$ out of $- 30$ .

$2 x^{2} + 2 (- 2 x) + 2 \cdot - 15 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 2 x)$ .

$2 (x^{2} - 2 x) + 2 \cdot - 15 = 0$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot - 15$ .

$2 (x^{2} - 2 x - 15) = 0$

Factor.

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$2 ((x - 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$2 (x - 5) (x + 3) = 0$

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

$x - 5 = 0$

$x + 3 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $2 (x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Solve for x 2x^2+10=4x+40

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