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

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Answer 1 · 2016-09-07T13:29:07

$x = \frac{3 - \sqrt{26}}{5}, \frac{3 + \sqrt{26}}{5}$ $x = (3 - sqrt(26)) / 5, (3 + sqrt(26)) / (5)$

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

$\Rightarrow x = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (5) (7)}}{2 (5)}$ $=> x = (- (- 6) pm sqrt((- 6)^(2) - 4 (5) (7))) / (2 (5))$

$\Rightarrow x = \frac{6 \pm \sqrt{36 - 140}}{10}$ $=> x = (6 pm sqrt(36 - 140)) / (10)$

$\Rightarrow x = \frac{6 \pm \sqrt{- 104}}{10}$ $=> x = (6 pm sqrt(- 104)) / (10)$

$\Rightarrow x = \frac{6 \pm \sqrt{104} i}{10}$ $=> x = (6 pm sqrt(104) i) / (10)$

$\Rightarrow x = \frac{6 \pm 2 \sqrt{26}}{10}$ $=> x = (6 pm 2 sqrt(26)) / (10)$

$\Rightarrow x = \frac{3 \pm \sqrt{26}}{5}$ $=> x = (3 pm sqrt(26)) / 5$

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

How do you solve 5x^2 -6x +7 = 05x2−6x+7=05x^2 -6x +7 = 0 using the quadratic formula?