How do you find the roots, real and imaginary, of $y=4x^2-x+12$ using the quadratic formula?

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Answer 1 · 2015-12-31T06:58:03

$x=(1+-isqrt191)/8$

For any quadratic equation $y=ax^2+bx+c$ , the roots are found through the formula

$x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case, when $y=4x^2-x+12$ ,

${(a=4),(b=-1),(c=12):}$

Plug these into the quadratic equation.

$x=(-(-1)+-sqrt((-1)^2-(4xx4xx12)))/(2xx4)$

$x=(1+-sqrt(1-192))/8$

$x=(1+-sqrt(-191))/8$

$x=(1+-isqrt191)/8$

The two solutions of this function are imaginary. The solution has no real solutions and will never cross the $x$ -axis.

graph{4x^2-x+12 [-20, 20, -10, 42.6]}

How do you find the roots, real and imaginary, of y=4x^2-x+12y=4x^2-x+12 using the quadratic formula?