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

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Answer 1 · 2016-11-15T03:38:18

$x=7.65 and x=-0.65$

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

$x=[-b+-sqrt(b^2-4ac)]/(2a)$

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

$x=[-(-7)+-sqrt((-7)^2-4xx1xx(-5))]/(2xx1)$

$x=[7+-sqrt(49+20)]/2$

$x=[7+-sqrt(69)]/2$

$x=[7+sqrt(69)]/2 and x=[7-sqrt(69)]/2$

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

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