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

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Answer 1 · 2016-11-15T03:38:18

$x = 7.65 and x = - 0.65$ $x=7.65 and x=-0.65$

$x^{2} - 7 x - 5 = 0$ $x^2-7x-5=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=[-b+-sqrt(b^2-4ac)]/(2a)$

Values given $a = 1, b = - 7, c = - 5$ $a=1, b=-7, c=-5$

$x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 1 \times (- 5)}}{2 \times 1}$ $x=[-(-7)+-sqrt((-7)^2-4xx1xx(-5))]/(2xx1)$

$x = \frac{7 \pm \sqrt{49 + 20}}{2}$ $x=[7+-sqrt(49+20)]/2$

$x = \frac{7 \pm \sqrt{69}}{2}$ $x=[7+-sqrt(69)]/2$

$x = \frac{7 + \sqrt{69}}{2} and x = \frac{7 - \sqrt{69}}{2}$ $x=[7+sqrt(69)]/2 and x=[7-sqrt(69)]/2$

$:.x=7.65 and x=-0.65$

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