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

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Answer 1 · 2017-08-24T14:26:37

$x=(1+isqrt(11))/4,(1-isqrt(11))/4$

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

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

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

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 $(1+isqrt(11))/4$ and $(1-isqrt(11))/4$ .

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