I have a Quadratic Equation that I input into the Quadratic Formula to solve for y.
$$2xy^{2}+8xy-z=0$$
$$y=\frac{-8x\pm \sqrt{(8x)^2 - 4(2x)(-z)}}{2(2x)}$$
I've been told factoring will take me to:
$$y=\sqrt{\frac{{z}}{2x}+4} -2$$
I'm not sure how they got from point A to point B.
Any help?
The numerator and denominator have been divided by $4x$, which becomes $16x^2$ inside the square root. Since $4x$ is the denominator, it disappears and we are left with $$y=-8x/4x\pm\sqrt{64x^2/16x^2+8xz/16x^2}=\sqrt{z/2x+4}-2$$ Clearly this only works if $x\ne0$.