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

Mar 7, 2016

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

#### Explanation:

The quadratic formula is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The a, b, and c refers to a, b, and c in standard form which is $a {x}^{2} + b x + c$. So given the equation $y = 7 {x}^{2} - 6 x + 14 , a = 7 , b = - 6 , c = 14$

Now we begin the process of plugging into the formula.

$\frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(7\right) \left(14\right)}}{2 \cdot \left(7\right)}$
$\frac{6 \pm \sqrt{36 - 392}}{14}$

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.