x2 + 12x + 27 = (x + 9)(x + 3)
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
x2 + 12x - 5 = 0x2 + 12x - 5 + 5 = 0 + 5x2 + 12x = 5x2 + 12x + (12/2)2 = 5 + (12/2)2x2 + 12x + (6)2 = 5 + 36(x + 6)2 = 41 (square root of both sides)x + 6 = ± √41x + 6 - 6 = -6 ± √41x = -6 ± √41x ≈ -6 + 6.4 = -0.4x ≈ -6 - 6.4 = -12.4
72 = x2 + x2 x = 6 x2 = 36 36+36=72
x2+12x+11=0 (x+11)(x+1)=0 x=-11 or x=-1
x2-12x+36 = (x-6)2
x2+12x+36=0 2x+12x+36=0 14x+36=0 14x=-36 x=-36/14 x=-2 4/7
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
=>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
x=5/2*sqr(3)
x2+12x+28Improved Answer:--x2+12x+28 = (-x+14)(x+2) when factored
x2+12x+5 = 3 x2+12x+5-3 = 0 x2+12x+2 = 0
Let's first reformat it: x2 + y2 - 12x - 8y - 43 = 0 x2 - 12x + 36 + y2 - 8y + 16 = 43 + 36 + 16 (x - 6)2 + (y - 4)2 = 95 So this equation defines a circle, with a center point at (6, 4) and a perimeter (i.e. circumference) of √95.
6.(x2 + 12x + 36) = 6.(x + 6)(x + 6)
Assuming the missing signs are pluses, that factors to (x + 6)(x + 6)
x2 + 12x - 11 = x2 + 12x + (12/2)2 - (12/2)2 - 11 = (x + 6)2 - 47
x2 - 12x3 = x2(1 - 12x)