SOLUTION: Solve the equation by completing the square and applying the square root property. Write imaginary solutions in the form a+bį. So far I have... -3y-10=-y^2 y^2-3y=10 Then

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve the equation by completing the square and applying the square root property. Write imaginary solutions in the form a+bį. So far I have... -3y-10=-y^2 y^2-3y=10 Then       Log On


   

Question 1059705: Solve the equation by completing the square and applying the square root property. Write imaginary solutions in the form a+bį.
So far I have...
-3y-10=-y^2
y^2-3y=10
Then what?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+y%5E2+-+3y+=+10+
+y%5E2+-+3y+%2B+%28+3%2F2+%29%5E2+=+10+%2B+%28+3%2F2+%29%5E2+
+y%5E2+-+3y+%2B+9%2F4+=+40%2F4+%2B+9%2F4+
+y%5E2+-+3y+%2B+9%2F4+=+49%2F4+
+%28+y+-+3%2F2+%29%5E2+=+%287%2F2+%29%5E2+
Take the square root of both sides
+y+-+3%2F2+=+7%2F2+
+y+=+10%2F2+
+y+=+5+
and, taking the negative square
root of +%287%2F2%29%5E2%0D%0A%7B%7B%7B+y+-+3%2F2+=+-7%2F2+
+y+=+-4%2F2+
+y+=+-2+
--------------------
check:
+y%5E2+-+3y+=+10+
+5%5E2+-+3%2A5+=+10+
+25+-+15+=+10+
+10+=+10+
-----------------------
+%28-2%29%5E2+-+3%2A%28-2%29+=+10+
+4+%2B+6+=+10+
+10+=+10+
OK