Question #11a9a

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Answer 1 · 2017-02-06T23:23:31

The discriminant is negative, so no real solutions exist.

The imaginary solutions are $x=-1.5+-(sqrt(-11))/2$

Since the equation is already in the standard form $ax^2+bx+c=0$ , the easiest way is to use the quadratic formula.

First, test the discriminant: $sqrt(b^2-4ac)$

This is negative: $sqrt(9-20) = sqrt(-11)$

This means there are no real solutions.

But, continuing (in case you are looking for imaginary solutions)

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

$x=((-3)+-sqrt(3^2-4(1)(5)))/(2(1))$

$x=((-3)+-sqrt(9-20))/2$

$x=-1.5+-(sqrt(-11))/2$