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

Question

6890 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-21T06:01:20

$x=(5+sqrt 29)/2$

$x=(5-sqrt29)/2$

$x^2-5x=1$

Subtract $1$ from both sides.

$x^2-5x-1=0$ is a quadratic equation in standard form, $ax^2+bx+c$ , where $a=1, b=-5, and c=-1$ .

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

Substitute the values for $a, b, and c$ into the formula.

$x=(-(-5)+-sqrt(-5^2-4*1*-1))/(2*1)$

Simplify.

$x=(5+-sqrt(25-(-4)))/2$

$x=(5+-sqrt(25+4))/2$

$x=(5+-sqrt29)/2$

Solve for $x$

$x=(5+sqrt 29)/2$

$x=(5-sqrt29)/2$

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