How do you solve $x^{2} - 3 x + 2 = 0$ $x^2-3x+2=0$ using the quadratic formula?

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How do you solve $x^{2} - 3 x + 2 = 0$ $x^2-3x+2=0$ using the quadratic formula?

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Answer 1 · 2018-05-02T15:05:37

$x_{1} = 2 or x_{2} = 1$ $x_1=2 or x_2=1$

Explanation:

$x^{2} + p x + q = 0$ $x^2+px+q=0$
$x_{1, 2} = - \frac{p}{2} \pm \sqrt{{(\frac{p}{2})}^{2} - q}$ $x_(1,2)=-p/2+-sqrt((p/2)^2-q)$
$p = - 3 and q = 2$ $p=-3 and q=2$
$x_{1, 2} = \frac{3}{2} \pm \sqrt{\frac{9}{4} - 2}$ $x_(1,2)=3/2+-sqrt(9/4-2)$
$x_{1} = \frac{3}{2} + \frac{1}{2} = 2 or x_{2} = \frac{3}{2} - \frac{1}{2} = 1$ $x_1=3/2+1/2=2 or x_2=3/2-1/2=1$

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How do you solve $x^{2} - 3 x + 2 = 0$ $x^2-3x+2=0$ using the quadratic formula?

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