# What is x value if f(x) = x^2 + 4x - 12

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f(x) = x^2 + 4x - 12

I assume that you need to solve for x values if f(x) = 0

==> x^2 + 4x - 12 = 0

Now we have two methods to solve:

1. Factor the equations:

==> ( x + 6) (x-2) = 0

==> x1= -6 and x2= 2

2. Use the forlula :

x= [-b +- sqrt(b^2 - 4ac)]/2a

==> x1= [ -4 + sqrt(16 - 4*-12)]/2

= [-4 + sqrt(64)]/2 = -4 + 8 /2 = 4/2 = 2

==> x2= (-4 - 8)/2 = -12/2 = -6

Then the answer is:

**x = { -6, 2}**

To solve for x , if x^2+4x-12 = 0.

We rewrite the given equation as : x^2+4x= 12.

We add 2^2 to the left side in or der that the left side is a perfect square:

x^2+4x+2^2 = 12+2^2

(x+2)^2 = 16.

We take square root of both sodes to solve for x:

x+2 = sqrt16 . Or x+2 = -sqrt16.

x+2 = 4 gives x = 4-2 = 2.

x+2 = -4 gives: x= -4-2 = -6.

Therefore x= 2 or x= -6 are solutions .

Tally:

We put x= 2 in x^2+4x-12 : 2^2+4*2-12 = 4+8-12 = 12-12 = 0 = RHS.

We put x= -6 in x^2+4x-12 : (-6)^2+4(-6) -12 = 36-24 -12 = 12-12 = 0 = RHS.