Solve by Factoring 2x^2+6x-80=0

$2 x^{2} + 6 x - 80 = 0$

Step 1

Factor $2$ out of $2 x^{2} + 6 x - 80$ .

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

$2 (x^{2}) + 6 x - 80 = 0$

Factor $2$ out of $6 x$ .

$2 (x^{2}) + 2 (3 x) - 80 = 0$

Factor $2$ out of $- 80$ .

$2 x^{2} + 2 (3 x) + 2 \cdot - 40 = 0$

Factor $2$ out of $2 x^{2} + 2 (3 x)$ .

$2 (x^{2} + 3 x) + 2 \cdot - 40 = 0$

Factor $2$ out of $2 (x^{2} + 3 x) + 2 \cdot - 40$ .

$2 (x^{2} + 3 x - 40) = 0$

Step 2

Factor.

Factor $x^{2} + 3 x - 40$ 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 $- 40$ and whose sum is $3$ .

$- 5, 8$

Write the factored form using these integers.

$2 ((x - 5) (x + 8)) = 0$

Remove unnecessary parentheses.

$2 (x - 5) (x + 8) = 0$

Step 3

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

$x - 5 = 0$

$x + 8 = 0$

Step 4

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.

$x = 5$

Step 5

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 6

The final solution is all the values that make $2 (x - 5) (x + 8) = 0$ true.

$x = 5, - 8$

Solve by Factoring 2x^2+6x-80=0

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