How do you find the zeros, real and imaginary, of #y=x^2+3x+7# using the quadratic formula?

1 Answer
Dec 7, 2015

Answer:

Plug the equation into the formula and solve for x. These will be your zeroes.

Explanation:

#y=ax^2+bx+c#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
plug in the values from the equation
#x=(-3+-sqrt(3^2-4(1)(7)))/(2(1))#
then solve for x...
#x=-3/2+-sqrt(-19)/2#

#x=(-3+-isqrt19)/2#
This gives 2 imaginary zeroes, but does not give any real zeroes, and if you graph the equation you will see that the equation does not have any.