# How do you solve the equation 3x^2-x+5=x^2+3x-14 by completing the square?

Feb 15, 2017

$2 {\left(x - 1\right)}^{2} + 17 = 0$

#### Explanation:

Combine like terms by moving all terms on the right to the left:
$\left(3 {x}^{2} - {x}^{2}\right) + \left(- x - 3 x\right) + \left(5 + 14\right) = 0$

Simplify: $2 {x}^{2} - 4 x + 19 = 0$

Complete the square:

1. Combine the x-terms & factoring: $2 \left({x}^{2} - 2 x\right) + 19 = 0$
2. Half the x-term $- 2 x : \frac{1}{2} \left(- 2\right) = - 1$, Used in ${\left(x - 1\right)}^{2}$
3. Square the halved x-term from step 2: ${\left(- 1\right)}^{2} = 1$, and multiply by $2$, the factored number in front of the square.
4. Complete the square & subtract the added term: $2 {\left(x - 1\right)}^{2} + 19 - 2 \left(1\right) = 0$
5. Simplify: $2 {\left(x - 1\right)}^{2} + 17 = 0$