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

1 Answer
Feb 16, 2016

Answer:

Solution is #(-1+isqrt15)/2# and #(-1-isqrt15)/2#

Explanation:

For an equation #ax^2+bx+c=0#, quadratic formula gives the solution as #(-b+-sqrt(b^2-4ac))/(2a)#.

As in the given equation #x^2+x+4=0#, #a=1, b=1# and #c=4#

Solution is given by #(-1+-sqrt(1^2-4*1.4))/(2*1)# or

#(-1+-sqrt(-15))/2# or #(-1+-isqrt15)/2#

Hence solution is #(-1+isqrt15)/2# and #(-1-isqrt15)/2#