Equate it to zero (0) , and you have a quadratic equation that can be factotred. Hence x^2 + 12x + 36 = 0 (x + 6)(x + 6) = 0 Hence x = -6
x2 + 12x + 32 = (x + 8)(x + 4)
x + 12x + 20 = (x + 10)(x + 2)
Add? I will show you the process. X^2 + 12X = 16 halve the linear term (12), square it and ADD it to both sides X^2 + 12X + 36 = 16 + 36 factor the right side; gather terms right (X + 6)^2 = 52 (X + 6)^2 - 52 = 0 (-6,-52) vertex of function As you see 36 was added to both sides
x2 + 13x + 36 = (x + 9)(x + 4)
x2-12x+36 = (x-6)2
x=5/2*sqr(3)
x2+12x+5 = 3 x2+12x+5-3 = 0 x2+12x+2 = 0
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
x2 - 12x + 36 = (x - 6)2
168
x2 + 12x - 64 = 0 ∴ (x + 16)(x - 4) = 0 ∴ x ∈ {4, -16}
It is -13.
x2+12x+28Improved Answer:--x2+12x+28 = (-x+14)(x+2) when factored
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.
x2 + 12x + 27 = (x + 9)(x + 3)