Question

How do you solve this?

Answer 1

$x^2-3x-4=0\\\\x^2-2x\cdot\frac{3}{2}=4\ \ \ /+(\frac{3}{2})^2\\\\x^2-2x\cdot\frac{3}{2}+(\frac{3}{2})^2=4+(\frac{3}{2})^2\\\\(x-\frac{3}{2})^2=4+\frac{9}{4}\\\\(x-\frac{3}{2})^2=\frac{16}{4}+\frac{9}{4}\\\\(x-\frac{3}{2})^2=\frac{25}{4}\\\\x-\frac{3}{2}=\pm\sqrt\frac{25}{4}$

$x-\frac{3}{2}=\pm\frac{\sqrt{25}}{\sqrt4}\\\\x-\frac{3}{2}=\pm\frac{5}{2}\\\\x-\frac{3}{2}=-\frac{5}{2}\ or\ x-\frac{3}{2}=\frac{5}{2}\\\\x=-\frac{5}{2}+\frac{3}{2}\ or\ x=\frac{5}{2}+\frac{3}{2}\\\\x=-\frac{2}{2}\ or\ x=\frac{8}{2}\\\\x=-1\ or\ x=4$

Answer 2

Why not simply factor the left side ?

(x - 4) times (x + 1) = 0

This equation is true if either factor is zero.

(x - 4) = 0
x = 4

(x + 1) = 0
x = -1