Question

How do you solve this?

Answer 1

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

Answer 2

`x^2-3x-4=0\\\\x^2-2x\cdot(3)/(2)=4\ \ \ /+((3)/(2))^2\\\\x^2-2x\cdot(3)/(2)+((3)/(2))^2=4+((3)/(2))^2\\\\(x-(3)/(2))^2=4+(9)/(4)\\\\(x-(3)/(2))^2=(16)/(4)+(9)/(4)\\\\(x-(3)/(2))^2=(25)/(4)\\\\x-(3)/(2)=\pm\sqrt(25)/(4)`


`x-(3)/(2)=\pm(√(25))/(\sqrt4)\\\\x-(3)/(2)=\pm(5)/(2)\\\\x-(3)/(2)=-(5)/(2)\ or\ x-(3)/(2)=(5)/(2)\\\\x=-(5)/(2)+(3)/(2)\ or\ x=(5)/(2)+(3)/(2)\\\\x=-(2)/(2)\ or\ x=(8)/(2)\\\\x=-1\ or\ x=4`