x2+12x+36=0 2x+12x+36=0 14x+36=0 14x=-36 x=-36/14 x=-2 4/7

This is not an inequality. This is an equation.12x+7=43 12x=36 x=3

x^2 + 12x = 5 x^2 + 12x + __ = 5 + __ x^2 + 12x + 36 = 5 + 36 (The 36 I got from taking half of 12 and then squaring that number (which is 6).) (x + 6)^2 = 41 x + 6 = (the square root of 41) x = (the square root of 41) - 6. Hope that helped :)

=>2x - 12x + 36 = 0 =>10x = 36=>x = 3.6Another contributor's answer:This is a quadratic equation question:x2-12x+36 = 0When factorised:(x-6)(x-6) = 0

Depending on whether or not there are implied brackets: x + 62 = x + 36 (x + 6)2 = x2 + 12x + 36

The graph is a circle with a radius of 6, centered at the origin.

6 squared is 36. 36 plus 3 is 39

36.1. You take the coefficient of x : = 122. Halve it : = 63. Then square it : = 364. Add it.This gives x2 + 12x + 36 = (x + 6)2The above method only works if the coefficient of x2 is 1. If it is not then the processs is slightly more complicated.

It is: 36 plus 121 = 157

36. 02 = 0 62 = 36

We have 3x+9y=36, or equivalently 12x+36y=144. We also have 12x+4y=18, so subtracting these we have 32y=126 or y=63/16. Substituting this into our original equation leaves 3x=36-9*63/16 or x = 12-3*63/16 = 3/16. So: x=3/16, y=63/16

If by d2 you mean d squared then it is non linear

36. 4 squared is 16...16 + 20 - = 36.

Because 6 squared equals 36 and 36 terminates, or ends, which makes it rational.

If you're looking to solve for x, you can do so as follows: x2 + 12x - 6 = 0 x2 + 12x + 36 = 42 (x + 6)2 = 42 x + 6 = ± √42 x = -6 ± √42

x^2 + 12x + 36 = 0(x + 6)(x + 6) = 0(x + 6)^2 = 0√(x + 6)^2 = ± √0x + 6 = 0x + 6 - 6 = 0 - 6x = 6

It is a quadratic equation and its solutions are: x = 4 or x = -9

Add 9x to each side: 12x + 11 = -25Subtract 11 from each side: 12x = -36Divide each side by 12: x = -3

The expression (12x + 36)/48x can be simplified by factoring out 12/12 from the expressions. This leaves (x + 3)/4x.

y - 8 = x^2 + 12x + 36 y - 8 = (x + 6)^2 y = (x + 6)^2 + 8 or f(x) = (x + 6)^2 + 8 So -h = 6 so that h = -6, and k = 8

The GCF of 12^2 + 18 - 36 is 6.

If: x^2+y^2 = 12x-10y-12 Then: x^2+y^2-12x+10y = -12 Completing the squares: (x-6)^2+(y+5)^2 -36-25 = -12 So: (x-6)^2+(y+5)^2 = 49 Therefore the centre of the circle is at (6, -5) and its radius is 7

x2 - 12x + 36 = (x - 6)2

6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.

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