Solving by using completing the square?

Question

x2+3x-7=0

3x2+3x-7=0

Answer 1

for the first one:
add seven
x^2+3x=7
now add to both sides the square of half of three. half of three is 1.5 the square of that is 2.25
x^2+3x+2.25=9.25
(x+1.5)^2=9.25
x+1.5=3.04 (approx)
x=1.54 (approx)

same for the second one, but after adding seven to both sides you divide both sides by three then do the rest the same as the first

Answer 2

Did you mean x^2 + 3x - 7 = 0 and 3x^2 + 3x - 7 = 0?

Each of them is a quadratic equation and can be solved either by factorisation or by using the quadratic formula. Both of them cannot be true for the same x.

Answer 3

x^2+3x-7=0
let y=x^2+3x-7
y=(x+3/2)^2-7-(3/2)^2
=(x+3/2)^2-7-9/4
=(x+3/2)^2-37/4
let y=3x^2+3x-7
put 3 outside brackets as the coefficient of x^2 must always be 1
therefore,
y=3(x^2+x-7/3)
=3[(x+1/2)^2-7/3-(1/2)^2]
=3[(x+1/2)^2-7/3-1/4]
=3[(x+1/2)^2-31/12]
remove big square brackets
=3(x+1/2)^2-31/4

(x+3/2)^2-37/4=0
3(x+1/2)^2-31/4=0
therefore (x+3/2)^2-37/4=3(x+1/2)^2-31/4
(x+3/2)^2-3(x+1/2)^2=-31/4+37/4
(x+3/2)^2-3(x+1/2)^2=3/2......(/3)
(x+3/2)^2-(x+1/2)^2=1/2

Answer 4

This page has an example: http://www.purplemath.com/modules/sqrquad.htm