Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-6x+5=0$$
To solve the equation, factor $x^{2}-6x+5$ 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.
$$a+b=-6$$ $$ab=5$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-5$$ $$b=-1$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-5\right)\left(x-1\right)$$
To find equation solutions, solve $x-5=0$ and $x-1=0$.
$$x=5$$ $$x=1$$
Steps Using Factoring By Grouping
Subtract $6x$ from both sides.
$$x^{2}+5-6x=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}-6x+5=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-6$$ $$ab=1\times 5=5$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-5$$ $$b=-1$$
Rewrite $x^{2}-6x+5$ as $\left(x^{2}-5x\right)+\left(-x+5\right)$.
$$\left(x^{2}-5x\right)+\left(-x+5\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(x-5\right)-\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(x-1\right)$$
To find equation solutions, solve $x-5=0$ and $x-1=0$.
$$x=5$$ $$x=1$$
Steps Using the Quadratic Formula
Subtract $6x$ from both sides.
$$x^{2}+5-6x=0$$
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.
$$x^{2}-6x+5=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-6$ for $b$, and $5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{6±4}{2}$ when $±$ is plus. Add $6$ to $4$.
$$x=\frac{10}{2}$$
Divide $10$ by $2$.
$$x=5$$
Now solve the equation $x=\frac{6±4}{2}$ when $±$ is minus. Subtract $4$ from $6$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
The equation is now solved.
$$x=5$$ $$x=1$$
Steps for Completing the Square
Subtract $6x$ from both sides.
$$x^{2}+5-6x=0$$
Subtract $5$ from both sides. Anything subtracted from zero gives its negation.
$$x^{2}-6x=-5$$
Divide $-6$, the coefficient of the $x$ term, by $2$ to get $-3$. Then add the square of $-3$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
