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#.