How do you solve $4x^2-7x+1=0$ using the quadratic formula?

Question

5346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T00:21:14

$x=~1.6,~0.16$

From the equation in standard form ( $ax^2+bx+c=0$ ), take the a, b, and c terms and sub them into the quadratic formula:

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

$x=(-(-7)+-sqrt(7^2-4(4)(1)))/(2*4)$

$x=(7+-sqrt(33))/8$

Now solve for the positive and negative roots.

Positive:

$x=(7+sqrt(33))/8$
$x=~1.6$

Negative:

$x=(7-sqrt(33))/8$
$x=~0.16$

How do you solve 4x^2-7x+1=04x^2-7x+1=0 using the quadratic formula?