x^2-4x-3=0?

Question

solve algebraically

Answer 1

A filthy little thing!
It looks like it is factorable by eye-balling it, but no. We shall have to use the quadratic formula.
x^2-4x-3=0
a=+1
b= -4
c=-3

x={-b+/-rt(b^2-4ac)} /2a
x={-(-4)+/-rt(-4^2-4(1)(-3)} /2
x={4+/-rt(16+12)} /2
x={4+/-rt28} /2
x=4+/-2rt7} /2
x=2(2+/-rt7)/2
x=2+rt7 or x=2-rt7....in decimal form,
x=2+2.65 or 2-2.65
x=4.65 or -0.65

Substituting both values back into x^2-4x-3=0 gives
0=0 in both cases. The answers are correct

Answer 2

x^2-4x-3=0
x^2-4x-3+3=0+3
x^2-4x=3
4q pi^poop
.

Answer 3

x^2-4x-3=0
x = [4 +/- sqrt(4^2 -4(1)(-3)]/2
x = [4 +/- sqrt(28)]/2
x = [4 +/- 2sqrt(7)]/2
x = 2 +/- sqrt(7)