How do you find the zeros, real and imaginary, of #y= 7x^2-6x+14# using the quadratic formula?

Question

Savannah Anderson · Accepted Answer

By plugging in the coefficients and constant as a, b, and c into the quadratic formula.

The quadratic formula is #(-b+-sqrt(b^2-4ac))/(2a)# The a, b, and c refers to a, b, and c in standard form which is #ax^2+bx+c#. So given the equation #y=7x^2-6x+14, a=7, b=-6, c=14# Now we begin the process of plugging into the formula. #(-(-6)+-sqrt((-6)^2-4(7)(14)))/(2*(7))# #(6+-sqrt(36-392))/14# Now at this point, I can stop because I know all the answers are imaginary, we know this because under the square root, there will be a negative number when we take 392 from 36. Because of this, the answers are imaginary.

Ethan Adkins · Answer

undefined To find the zeros of the quadratic equation $ y = 7x^2 - 6x + 14 $ using the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 7 $, $ b = -6 $, and $ c = 14 $.

Substitute the values into the quadratic formula:

$$ x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 7 \cdot 14}}}}{{2 \cdot 7}} $$

$$ x = \frac{{6 \pm \sqrt{{36 - 392}}}}{{14}} $$

$$ x = \frac{{6 \pm \sqrt{{-356}}}}{{14}} $$

$$ x = \frac{{6 \pm \sqrt{{356i}}}}{{14}} $$

$$ x = \frac{{6 \pm 2\sqrt{{89i}}}}{{14}} $$

So the zeros of the equation are:

$$ x = \frac{{3 \pm \sqrt{{89i}}}}{{7}} $$