Questions on Algebra: Expressions involving variables, substitution answered by real tutors!

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Question 153536: Please help me solve the following problem:
Solve: x^2 + 3x - 4 = 0
: Please help me solve the following problem:
Solve: x^2 + 3x - 4 = 0

Answer by stanbon(19020) About Me  (Show Source):
You can put this solution on YOUR website!
Solve: x^2 + 3x - 4 = 0
Factor to get:
(x+4)(x-1) = 0
x = -4 or x = 1
=====================
Cheers,
Stan H.
Question 153536: Please help me solve the following problem:
Solve: x^2 + 3x - 4 = 0
: Please help me solve the following problem:
Solve: x^2 + 3x - 4 = 0

Answer by josmiceli(2035) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + 3x - 4
Solving this means finding the roots, or finding
the values of x that make x^2 + 3x - 4 = 0
true.
If I just say, off the top of my head, x = 2 for
a solution, then I'm saying 2^2 + 3*2 - 4 = 0 and
4 + 6 - 4 = 0
10 - 4 = 0
6 = 0
That's not true, so x = 2 is not a solution
One way to solve this is by completing the square.
First add 4 to both sides
x^2 + 3x = 4
Now take one-half of the coefficient of the x term,
square it, and add it to both sides, like this:
x^2 + 3x + (3/2)^2 = 4 + (3/2)^2
x^2 + 3x + (9/4) = (16/4) + (9/4)
x^2 + 3x + (9/4) = 25/4
This is the same as
(x + (3/2))^2 = 25/4
Now take the square root of both sides
x + (3/2) = 5/2
and also
x + (3/2) = -(5/2) (since 25/4 has a + and a - square root)
Subtract 3/2 from both sides
x = (5/2) - (3/2)
x = 1
And also
x = -(5/2) - (3/2)
x = -4
So, the solutions are x = 1 and x = -4
Plug these values into the equation to see if they
are roots, for instance:
(-4)^2 + 3*(-4) - 4 = 0
16 - 12 - 4 = 0
16 - 16 = 0
0 = 0
So, x = -4 is a solution