# How do you find the zeros, real and imaginary, of y=4x^2-2x+3 using the quadratic formula?

Aug 24, 2017

$x = \frac{1 + i \sqrt{11}}{4} , \frac{1 - i \sqrt{11}}{4}$

#### Explanation:

Let's find the values for variables a, b, and c.

$y = 4 {x}^{2} - 2 x + 3$

$a = 4 , b = - 2 , c = 3$

Let's plug these numbers into the equation.

x=(-b±sqrt(b^2-4ac))/(2a)=(-(-2)±sqrt((-2)^2-4(4)(3)))/(2(4))

Now we can solve for the zeros of the equation.

x=(2±sqrt(4-48))/8 ->

x=(2±sqrt(-44))/8 ->

x=(2±2isqrt(11))/8 ->

x=(1±isqrt(11))/4

Your zeroes will be imaginary. The zeroes are $\frac{1 + i \sqrt{11}}{4}$ and $\frac{1 - i \sqrt{11}}{4}$.