To solve the equation, factor $x^{2}-10x+25$ 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.
$$a+b=-10$$ $$ab=25$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $25$.
$$-1,-25$$ $$-5,-5$$
Calculate the sum for each pair.
$$-1-25=-26$$ $$-5-5=-10$$
The solution is the pair that gives sum $-10$.
$$a=-5$$ $$b=-5$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-5\right)\left(x-5\right)$$
Rewrite as a binomial square.
$$\left(x-5\right)^{2}$$
To find equation solution, solve $x-5=0$.
$$x=5$$
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 $x^{2}+ax+bx+25$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=1\times 25=25$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $25$.
$$-1,-25$$ $$-5,-5$$
Calculate the sum for each pair.
$$-1-25=-26$$ $$-5-5=-10$$
The solution is the pair that gives sum $-10$.
$$a=-5$$ $$b=-5$$
Rewrite $x^{2}-10x+25$ as $\left(x^{2}-5x\right)+\left(-5x+25\right)$.
$$\left(x^{2}-5x\right)+\left(-5x+25\right)$$
Factor out $x$ in the first and $-5$ in the second group.
$$x\left(x-5\right)-5\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(x-5\right)$$
Rewrite as a binomial square.
$$\left(x-5\right)^{2}$$
To find equation solution, solve $x-5=0$.
$$x=5$$
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.
$$x^{2}-10x+25=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-10$ for $b$, and $25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$x^{2}-10x+25=0$$
Factor $x^{2}-10x+25$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-5\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$
Simplify.
$$x-5=0$$ $$x-5=0$$
Add $5$ to both sides of the equation.
$$x=5$$ $$x=5$$
The equation is now solved. Solutions are the same.
