# How do you solve x^2 - 6x + 6 = 0 using the quadratic formula?

Jul 6, 2016

${x}_{1} = 4.73$ (3sf) ${x}_{2} = 1.27$ (3sf)

#### Explanation:

The quadratic formula is: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, Therefore the first step is to identify $a , b$ & $c$. We need to ensure that the equation is $= 0$ and in this case that is true.

Then we are able to obtain the values to substitute into the formula; the general quadratic formula layout is $a {x}^{2} + b x + c$ therefore we look at your equation: ${x}^{2} - 6 x + 6$. The value of $a$ has to be $1$ since ${x}^{2}$ is multiplied by $1$. Then $b = - 6$, maintaining the negative is essential to the calculation (the sign will also stay with the value that is next in the equation). Then $c = 6$.

We now substitute $a = 1$, $b = - 6$ and $c = 6$ into the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ $= \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(1\right) \left(6\right)}}{2 \left(1\right)}$

Now that we have the values substituted into the equation we first solve the inside of the square root. So,

$x = \frac{- \left(- 6\right) \pm \sqrt{12}}{2 \left(1\right)}$

Since the value under the root (discriminant) is positive we know that there will be two real solutions for $x$.

Separate the two $x$ values so that we solve for the positive root first.

${x}_{1} = \frac{- \left(- 6\right) + \sqrt{12}}{2 \left(1\right)}$ = $4.73$ (3 sf)

Now solve the second $x$ value by using the negative root.

${x}_{2} = \frac{- \left(- 6\right) - \sqrt{12}}{2 \left(1\right)}$ = $1.27$ (3 sf)