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

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Answer 1 · 2016-09-23T13:28:16

$x = (5 pm sqrt(13)) / (2)$

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

$=> x = (- (- 5) pm sqrt((- 5)^(2) - 4 (1) (3))) / (2 (1))$

$=> x = (5 pm sqrt(25 - 12)) / (2)$

$=> x = (5 pm sqrt(13)) / (2)$

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

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