SOLUTION: (3y+2)^2+400=0

Question 346505: (3y+2)^2+400=0
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

If we subtract 400 from each side we get:

At this point, if we're paying attention to what the equation is saying, we can figure out that there is no solution (at least within the set of Real numbers) to this equation. The equation is saying that something squared equals a negative number. Within the set of Real numbers squaring a number always results in either a positive number or zero. So squaring a number and getting -400 is not possible.

If you are working with the set of Complex numbers, then this equation is possible. So I will continue a solution with the understanding that the answer will be a complex number.

We can proceed by finding the square root(s) of each side:

Simplifying this:

(Note: The "0" is used here because Algebra.com's formula software will not let me use "+-" without it.)

Now we can solve for y. Add -2 to each side:

Multiply each side by 1/3:

These are the complex number solutions to your equation.
RELATED QUESTIONS
y2-3y+2=0 (answered by brandonpark2889)

2(0)+3y=12 (answered by rfer)

[[[3y^2-2y-4=0}}} (answered by robvandam)

{{{ 3y^2-7y-5=0... (answered by jim_thompson5910,checkley71)

{{{3y^2+7y+1=0}}} (answered by drj)

solve 3y^2+4y-2<=... (answered by stanbon)

2y^-2 -3y^-1 -2 =0 (answered by troymuga)

7x^2+3y^2-21=0 (answered by lwsshak3)

3y-2(3y+2)=8 (answered by rfer)