Solve for q 2q^2-13q+20=0

$2 q^{2} - 13 q + 20 = 0$

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 $a \cdot c = 2 \cdot 20 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 q$ .

$2 q^{2} - 13 q + 20 = 0$

Rewrite $- 13$ as $- 5$ plus $- 8$

$2 q^{2} + (- 5 - 8) q + 20 = 0$

Apply the distributive property.

$2 q^{2} - 5 q - 8 q + 20 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 q^{2} - 5 q) - 8 q + 20 = 0$

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

$q (2 q - 5) - 4 (2 q - 5) = 0$

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

$(2 q - 5) (q - 4) = 0$

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

$2 q - 5 = 0$

$q - 4 = 0$

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

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

$2 q - 5 = 0$

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

Add $5$ to both sides of the equation.

$2 q = 5$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $q$ by $1$ .

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

Set $q - 4$ equal to $0$ and solve for $q$ .

Set $q - 4$ equal to $0$ .

$q - 4 = 0$

Add $4$ to both sides of the equation.

$q = 4$

The final solution is all the values that make $(2 q - 5) (q - 4) = 0$ true.

$q = \frac{5}{2}, 4$

The result can be shown in multiple forms.

Exact Form:

$q = \frac{5}{2}, 4$

Decimal Form:

$q = 2.5, 4$

Mixed Number Form:

$q = 2 \frac{1}{2}, 4$

Solve for q 2q^2-13q+20=0

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