solve 2x^2+3x+8=0 and express the solutions in a+bi form.

Let's use the quadratic formula to solve for x and express those solutions in a+bi form.

x = [-b + or - sqrt(b^2 - 4ac)]/2a

Note: sqrt = square root.

a = 2, b = 3, and c = 8 from your problem.

Therefore:
x = [-3 + or - sqrt(3^2 - 4*2*8)]/2*2
x = [-3 + or - sqrt(-55)]4
x = [-3 + or - i sqrt(55)]/4

Solutions:
x = [-3 + i sqrt(55)]/4 = -3/4 + [sqrt(55)/4]i
x = [-3 - i sqrt(55)]/4 = -3/4 - [sqrt(55)/4]i

I hope this will help.

Itm ay have taken almost 20 years and you are probably in your 30s-40s (and I'm not even joking but my actual dad was named "Jimmy" so this person might be him, who knows) but this finally helped! Thanks high school student from soon to be 20 years ago!! And hello future students that may come across this years later <3

(my insta is @jaineykeyes if you wanna know how I was like or what I could be doing at the time you find this)

My math was completely wrong by +5 years (you can tell I suck at math loll)

2007 is about to be 15 years ago so they should be in their 20s-30s now.

The solutions to the equation 2x^2 + 3x + 8 = 0, expressed in a+bi form, are:

x = (-3/4 + (sqrt(55)/4)i) and x = (-3/4 - (sqrt(55)/4)i)

So the solutions are (-3/4 + (sqrt(55)/4)i) and (-3/4 - (sqrt(55)/4)i).

To solve the quadratic equation 2x^2 + 3x + 8 = 0 and express the solutions in a+bi form, we can use the quadratic formula.

The quadratic formula is:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

First, let's identify the values of a, b, and c in the equation:

a = 2
b = 3
c = 8

Now we substitute these values into the quadratic formula:

x = [-3 ± sqrt(3^2 - 4 * 2 * 8)] / (2 * 2)

Simplifying further:

x = [-3 ± sqrt(9 - 64)] / 4
x = [-3 ± sqrt(-55)] / 4

Since the expression inside the square root is negative, we can rewrite it in terms of the imaginary unit i:

x = [-3 ± i * sqrt(55)] / 4

Therefore, the solutions in a+bi form are:

x = (-3 ± i * sqrt(55)) / 4

To write the solutions explicitly, we can separate the real and imaginary parts:

x1 = -3/4 + (sqrt(55)/4)i
x2 = -3/4 - (sqrt(55)/4)i

So the solutions to the equation 2x^2 + 3x + 8 = 0 in the form a+bi are:

x1 = -3/4 + (sqrt(55)/4)i
x2 = -3/4 - (sqrt(55)/4)i