How do you solve $5x^2 -6x +7 = 0$ using the quadratic formula?

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Answer 1 · 2016-09-07T13:29:07

$x = (3 - sqrt(26)) / 5, (3 + sqrt(26)) / (5)$

We have: $5 x^(2) - 6 x + 7 = 0$

$=> x = (- (- 6) pm sqrt((- 6)^(2) - 4 (5) (7))) / (2 (5))$

$=> x = (6 pm sqrt(36 - 140)) / (10)$

$=> x = (6 pm sqrt(- 104)) / (10)$

$=> x = (6 pm sqrt(104) i) / (10)$

$=> x = (6 pm 2 sqrt(26)) / (10)$

$=> x = (3 pm sqrt(26)) / 5$

Therefore, the solutions to the equation are $x = 3 - sqrt(26)$ and $x = 3 + sqrt(26)$ .

How do you solve 5x^2 -6x +7 = 05x^2 -6x +7 = 0 using the quadratic formula?