How do you solve $x^{2} - 5 x + 3 = 0$ $x^2-5x+3=0$ using the quadratic formula?

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How do you solve $x^{2} - 5 x + 3 = 0$ $x^2-5x+3=0$ using the quadratic formula?

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Answer 1 · 2016-09-23T13:28:16

$x = \frac{5 \pm \sqrt{13}}{2}$ $x = (5 pm sqrt(13)) / (2)$

Explanation:

We have: $x^{2} - 5 x + 3 = 0$ $x^(2) - 5 x + 3 = 0$

$\Rightarrow x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (1) (3)}}{2 (1)}$ $=> x = (- (- 5) pm sqrt((- 5)^(2) - 4 (1) (3))) / (2 (1))$

$\Rightarrow x = \frac{5 \pm \sqrt{25 - 12}}{2}$ $=> x = (5 pm sqrt(25 - 12)) / (2)$

$\Rightarrow x = \frac{5 \pm \sqrt{13}}{2}$ $=> x = (5 pm sqrt(13)) / (2)$

Therefore, the solutions to the equation are $x = \frac{5 - \sqrt{13}}{2}$ $x = (5 - sqrt(13)) / (2)$ and $x = \frac{5 + \sqrt{13}}{2}$ $x = (5 + sqrt(13)) / (2)$ .

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How do you solve $x^{2} - 5 x + 3 = 0$ $x^2-5x+3=0$ using the quadratic formula?

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Explanation:

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How do you solve $x^{2} - 5 x + 3 = 0$ $x^2-5x+3=0$ using the quadratic formula?