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

1 Answer
Alexander A.
Dec 7, 2015

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.

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