Question

How to solve the following equation? Or at least tell me how many real solutions it has:

Answer 1

\[4\sqrt{4-x ^{2}}+3x=-20\]

Answer 2

sorry ... second degree equation .. just separate the root on one side and square

Answer 3

Oh yeah..that didn't come to my mind. Thanks!

Answer 4

4 sqrt(4-x^2) = -20-3x
16( 4-x^2) = 400+120x+9x^2
64-16x^2 = 400 +120x +9x^2
25x^2+120x+336=0
b^2 -4ac = 120^2 - 4(25)(336) <0
therefore, no real solution