To solve the equation, factor $x^{2}+34x+289$ 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.
$$a+b=34$$ $$ab=289$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $289$.
$$1,289$$ $$17,17$$
Calculate the sum for each pair.
$$1+289=290$$ $$17+17=34$$
The solution is the pair that gives sum $34$.
$$a=17$$ $$b=17$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x+17\right)\left(x+17\right)$$
Rewrite as a binomial square.
$$\left(x+17\right)^{2}$$
To find equation solution, solve $x+17=0$.
$$x=-17$$
Steps Using Factoring By Grouping
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+289$. To find $a$ and $b$, set up a system to be solved.
$$a+b=34$$ $$ab=1\times 289=289$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $289$.
$$1,289$$ $$17,17$$
Calculate the sum for each pair.
$$1+289=290$$ $$17+17=34$$
The solution is the pair that gives sum $34$.
$$a=17$$ $$b=17$$
Rewrite $x^{2}+34x+289$ as $\left(x^{2}+17x\right)+\left(17x+289\right)$.
$$\left(x^{2}+17x\right)+\left(17x+289\right)$$
Factor out $x$ in the first and $17$ in the second group.
$$x\left(x+17\right)+17\left(x+17\right)$$
Factor out common term $x+17$ by using distributive property.
$$\left(x+17\right)\left(x+17\right)$$
Rewrite as a binomial square.
$$\left(x+17\right)^{2}$$
To find equation solution, solve $x+17=0$.
$$x=-17$$
Steps Using the Quadratic Formula
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.
$$x^{2}+34x+289=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $34$ for $b$, and $289$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-34±\sqrt{34^{2}-4\times 289}}{2}$$
Square $34$.
$$x=\frac{-34±\sqrt{1156-4\times 289}}{2}$$
Multiply $-4$ times $289$.
$$x=\frac{-34±\sqrt{1156-1156}}{2}$$
Add $1156$ to $-1156$.
$$x=\frac{-34±\sqrt{0}}{2}$$
Take the square root of $0$.
$$x=-\frac{34}{2}$$
Divide $-34$ by $2$.
$$x=-17$$
Steps for Completing the Square
Factor $x^{2}+34x+289$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+17\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x+17\right)^{2}}=\sqrt{0}$$
Simplify.
$$x+17=0$$ $$x+17=0$$
Subtract $17$ from both sides of the equation.
$$x=-17$$ $$x=-17$$
The equation is now solved. Solutions are the same.
