How do you solve #x^2 +5x-7=0#?

1 Answer
May 24, 2016

Answer:

#x = (-5+-sqrt(53))/2#

Explanation:

This quadratic equation is in the form #ax^2+bx+c = 0# with #a=1#, #b=5# and #c=-7#.

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 5^2-(4*1*-7) = 25+28 = 53#

So #Delta > 0#, but it is not a perfect square, so our quadratic equation has irrational roots.

We can find them using the quadratic formula:

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

#=(-b+-sqrt(Delta))/(2a)#

#=(-5+-sqrt(53))/2#