Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-5x-14=0$$
To solve the equation, factor $x^{2}-5x-14$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=-14$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-14$.
$$1,-14$$ $$2,-7$$
Calculate the sum for each pair.
$$1-14=-13$$ $$2-7=-5$$
The solution is the pair that gives sum $-5$.
$$a=-7$$ $$b=2$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-7\right)\left(x+2\right)$$
To find equation solutions, solve $x-7=0$ and $x+2=0$.
$$x=7$$ $$x=-2$$
Steps Using Factoring By Grouping
Subtract $5x$ from both sides.
$$x^{2}-14-5x=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-5x-14=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-14$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=1\left(-14\right)=-14$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-14$.
$$1,-14$$ $$2,-7$$
Calculate the sum for each pair.
$$1-14=-13$$ $$2-7=-5$$
The solution is the pair that gives sum $-5$.
$$a=-7$$ $$b=2$$
Rewrite $x^{2}-5x-14$ as $\left(x^{2}-7x\right)+\left(2x-14\right)$.
$$\left(x^{2}-7x\right)+\left(2x-14\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x-7\right)+2\left(x-7\right)$$
Factor out common term $x-7$ by using distributive property.
$$\left(x-7\right)\left(x+2\right)$$
To find equation solutions, solve $x-7=0$ and $x+2=0$.
$$x=7$$ $$x=-2$$
Steps Using the Quadratic Formula
Subtract $5x$ from both sides.
$$x^{2}-14-5x=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$x^{2}-5x-14=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-5$ for $b$, and $-14$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{5±9}{2}$ when $±$ is plus. Add $5$ to $9$.
$$x=\frac{14}{2}$$
Divide $14$ by $2$.
$$x=7$$
Now solve the equation $x=\frac{5±9}{2}$ when $±$ is minus. Subtract $9$ from $5$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
The equation is now solved.
$$x=7$$ $$x=-2$$
Steps for Completing the Square
Subtract $5x$ from both sides.
$$x^{2}-14-5x=0$$
Add $14$ to both sides. Anything plus zero gives itself.
$$x^{2}-5x=14$$
Divide $-5$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{2}$. Then add the square of $-\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
