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

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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#