Solve by Factoring 2x^2+9x-5=0

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

Factor by grouping.

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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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 9$ .

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Factor $9$ out of $9 x$ .

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

Rewrite $9$ as $- 1$ plus $10$

$2 x^{2} + (- 1 + 10) x - 5 = 0$

Apply the distributive property.

$2 x^{2} - 1 x + 10 x - 5 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 10 x - 5 = 0$

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

$x (2 x - 1) + 5 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 5) = 0$

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

$2 x - 1 = 0$

$x + 5 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$2 x - 1 = 0$

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(2 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{2}, - 5$

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