# What are the correct values of x in the equation 4x^2 = y when y = 144?

First, we pass the "4" that multiplies x to divide 144: x² = 144 / 4 = 36
Then, we have to pass the square of x to the other side, with its value inverted: x² = 36 >> $x = {36}^{\frac{1}{2}} = \sqrt{36} = \pm 6$ . So, the first value of X is +6, and the second is -6