How do you solve #2x^2 - 7x - 3 = 0 #?

1 Answer
Alan P.
Sep 9, 2015

#x=(7+-sqrt(73))/4#

Explanation:

An examination of the factors of #2# and #-3# reveals that there are no integer solutions.

Therefore, we will use the quadratic root formula
#color(white)("XXX")x= (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)("XXXXXXXXXXX")#for a quadratic in the general form #ax^2+bx+c=0#

In this case
#color(white)("XXX")x = (7+-sqrt(7^2-4(2)(-3)))/(2(2))#

#color(white)("XXXX")=(7+-sqrt(49+24))/4#

#color(white)("XXXX")=(7+-sqrt(73))/4#

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