Swap sides so that all variable terms are on the left hand side.
$$x^{2}+2x=24$$
Subtract $24$ from both sides.
$$x^{2}+2x-24=0$$
To solve the equation, factor $x^{2}+2x-24$ 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.
$$a+b=2$$ $$ab=-24$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-24$.
$$-1,24$$ $$-2,12$$ $$-3,8$$ $$-4,6$$
Calculate the sum for each pair.
$$-1+24=23$$ $$-2+12=10$$ $$-3+8=5$$ $$-4+6=2$$
The solution is the pair that gives sum $2$.
$$a=-4$$ $$b=6$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-4\right)\left(x+6\right)$$
To find equation solutions, solve $x-4=0$ and $x+6=0$.
$$x=4$$ $$x=-6$$
Steps Using Factoring By Grouping
Swap sides so that all variable terms are on the left hand side.
$$x^{2}+2x=24$$
Subtract $24$ from both sides.
$$x^{2}+2x-24=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-24$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-24\right)=-24$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-24$.
$$-1,24$$ $$-2,12$$ $$-3,8$$ $$-4,6$$
Calculate the sum for each pair.
$$-1+24=23$$ $$-2+12=10$$ $$-3+8=5$$ $$-4+6=2$$
The solution is the pair that gives sum $2$.
$$a=-4$$ $$b=6$$
Rewrite $x^{2}+2x-24$ as $\left(x^{2}-4x\right)+\left(6x-24\right)$.
$$\left(x^{2}-4x\right)+\left(6x-24\right)$$
Factor out $x$ in the first and $6$ in the second group.
$$x\left(x-4\right)+6\left(x-4\right)$$
Factor out common term $x-4$ by using distributive property.
$$\left(x-4\right)\left(x+6\right)$$
To find equation solutions, solve $x-4=0$ and $x+6=0$.
$$x=4$$ $$x=-6$$
Steps Using the Quadratic Formula
Swap sides so that all variable terms are on the left hand side.
$$x^{2}+2x=24$$
Subtract $24$ from both sides.
$$x^{2}+2x-24=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-24$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-2±\sqrt{2^{2}-4\left(-24\right)}}{2}$$
Square $2$.
$$x=\frac{-2±\sqrt{4-4\left(-24\right)}}{2}$$
Multiply $-4$ times $-24$.
$$x=\frac{-2±\sqrt{4+96}}{2}$$
Add $4$ to $96$.
$$x=\frac{-2±\sqrt{100}}{2}$$
Take the square root of $100$.
$$x=\frac{-2±10}{2}$$
Now solve the equation $x=\frac{-2±10}{2}$ when $±$ is plus. Add $-2$ to $10$.
$$x=\frac{8}{2}$$
Divide $8$ by $2$.
$$x=4$$
Now solve the equation $x=\frac{-2±10}{2}$ when $±$ is minus. Subtract $10$ from $-2$.
$$x=-\frac{12}{2}$$
Divide $-12$ by $2$.
$$x=-6$$
The equation is now solved.
$$x=4$$ $$x=-6$$
Steps for Completing the Square
Swap sides so that all variable terms are on the left hand side.
$$x^{2}+2x=24$$
Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+2x+1^{2}=24+1^{2}$$
Square $1$.
$$x^{2}+2x+1=24+1$$
Add $24$ to $1$.
$$x^{2}+2x+1=25$$
Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+1\right)^{2}=25$$
Take the square root of both sides of the equation.
