Solve for x 2x^2+x-15=0

$2 x^{2} + x - 15 = 0$

Step 1

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = 1$ .

Multiply by $1$ .

$2 x^{2} + 1 x - 15 = 0$

Rewrite $1$ as $- 5$ plus $6$

$2 x^{2} + (- 5 + 6) x - 15 = 0$

Apply the distributive property.

$2 x^{2} - 5 x + 6 x - 15 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 5 x) + 6 x - 15 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

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

Step 2

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

$2 x - 5 = 0$

$x + 3 = 0$

Step 3

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 4

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.

$x = - 3$

Step 5

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

$x = \frac{5}{2}, - 3$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2}, - 3$

Decimal Form:

$x = 2.5, - 3$

Mixed Number Form:

$x = 2 \frac{1}{2}, - 3$

Solve for x 2x^2+x-15=0

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