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

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Answer 1 · 2018-05-03T04:35:56

$x=(1-isqrt47)/8,$ $(1+isqrt47)/8$

The roots are imaginary.

$y=-4x^2+x-3$ is a quadratic equation in standard form:

$y=ax^2+x-3$ ,

where:

$a=-4$ , $b=1$ , $c=-3$

To find the roots, substitute $0$ for $y$ .

$0=-4x^2+x-3$

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

Plug in the known values.

$x=(-1+-sqrt(1^2-4*-4*-3))/(2*-4)$

Simplify.

$x=(-1+-sqrt(-47))/(-8)$

$x=(-1+-sqrt(47*-1))/(-8)$

Simplify $sqrt(-1)$ to $i$ .

$x=(-1+-isqrt47)/(-8)$

$47$ is a prime number so it can't be factored.

$x=(1+-isqrt47)/8$ $larr$ Two negatives make a positive.

Roots

$x=(1-isqrt47)/8,$ $(1+isqrt47)/8$

graph{y=-4x^2+x-3 [-15.02, 17, -15.25, 0.77]}

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