# How do you find the zeros, real and imaginary, of y=4x^2+4x+6 using the quadratic formula?

Jun 13, 2016

Zeros of $4 {x}^{2} + 4 x - 6$ are $\left(- \frac{1}{2} - \frac{\sqrt{7}}{2}\right)$ and $\left(- \frac{1}{2} + \frac{\sqrt{7}}{2}\right)$

#### Explanation:

In $4 {x}^{2} + 4 x - 6$, as the discriminant is ${4}^{2} - 4 \cdot 4 \cdot \left(- 6\right) = \left(16 + 96\right) = - 112$, is not the square of a rational number, we cannot factorize it by splitting middle term and for finding zeros will have to use quadratic formula. Further, as it is positive zeros will be irrational conjugates.

Zeros of quadratic trinomial $4 {x}^{2} + 4 x - 6$ as zeros of $a {x}^{2} + b x + c$ are given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence zeros of $4 {x}^{2} + 4 x - 6$ are

$\frac{- 4 \pm \sqrt{112}}{2 \times 4}$ or

$\frac{- 4 \pm 4 \sqrt{7}}{8}$ or

$- \frac{1}{2} - \frac{\sqrt{7}}{2}$ and $- \frac{1}{2} + \frac{\sqrt{7}}{2}$