# How do you solve x^2-7+12=0?

Jun 12, 2015

I am not sure about the question. I think it should be ${x}^{2} - 7 x + 12 = 0$.

#### Explanation:

Case 1 (I do not think it is the right one).
${x}^{2} - 7 + 12 = 0$
${x}^{2} = - 5$
$x = \pm \sqrt{5} \cdot i$
Where $i = \sqrt{- 1}$

Case 2
${x}^{2} - 7 x + 12 = 0$

${x}_{1 , 2} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2}$
${x}_{1} = 4$
${x}_{2} = 3$