How do you solve $x^2-5x-1 = 0$ using the quadratic formula?

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Answer 1 · 2016-04-14T15:29:38

$x_1=0,195$
$x_2=5,195$

$ax^2+bx+c=0$

$x_"1,2"=(-b±sqrt(b^2-4*a*c))/(2*a)" the quadratic formula"$

$x^2-5x-1=0$

$a=1" "b=-5" "c=-1$

$Delta=sqrt(b^2-4*a*c)$

$Delta=sqrt(25+4*1*1)$

$Delta=±5,39$

$x_1=(-b-Delta)/(2*a)$

$x_1=(5-5,39)/(2*1)$

$x_1=(0,39)/2$

$x_1=0,195$

$x_2=(-b+Delta)/(2*a)$

$x_2=(5+5,39)/2$

$x_2=(10,39)/2$

$x_2=5,195$

How do you solve x^2-5x-1 = 0x^2-5x-1 = 0 using the quadratic formula?