How do you solve $x^{2} - 7 + 12 = 0$ $x^2-7+12=0$ ?

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How do you solve $x^{2} - 7 + 12 = 0$ $x^2-7+12=0$ ?

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Answer 1 · 2015-06-12T18:22:36

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

Explanation:

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

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

Use the Quadratic Formula:

$x_{1, 2} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2}$ $x_(1,2)=(7+-sqrt(49-48))/2=(7+-1)/2$
Two solutions:
$x_{1} = 4$ $x_1=4$
$x_{2} = 3$ $x_2=3$

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How do you solve $x^{2} - 7 + 12 = 0$ $x^2-7+12=0$ ?

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How do you solve $x^{2} - 7 + 12 = 0$ $x^2-7+12=0$ ?