# How do you solve  x^2+6x-1=0?

Jun 14, 2015

Solve this equation by completing the square.

$x = \sqrt{10} - 3$

$x = - \sqrt{10} - 3$

#### Explanation:

Solve ${x}^{2} + 6 x - 1 = 0$

We are going to create a perfect square trinomial on the left side of the equation. A perfect square trinomial has the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$. Then we can factor the perfect square trinomial and solve for $x$.

Add the constant term $\left(1\right)$ to both sides of the equation.

${x}^{2} + 6 x = 1$

Divide the coefficient of the $x$ term by $2$, square the result, and add it to both sides of the equation.

$\frac{6}{2} = 3$; ${3}^{2} = 9$

${x}^{2} + 6 x + 9 = 10$

We now have a perfect square trinomial on the left side which can be factored to ${\left(x + 3\right)}^{2}$.

${\left(x + 3\right)}^{2} = 10$

Take the square root of both sides.

$x + 3 = \pm \sqrt{10}$

$x = \pm \sqrt{10} - 3$

$x = \sqrt{10} - 3$

$x = - \sqrt{10} - 3$