x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(2-r)e-rr
x2 + 4x = 1 x2 + 4x - 1 = 0 in the quadratic formula (b2+- Sqr(4ac))/2a, a = 1 b = 4 c = -1 so x = 8 +- i
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
No.
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(2-r)e-rr
x2 + 4x = 1 x2 + 4x - 1 = 0 in the quadratic formula (b2+- Sqr(4ac))/2a, a = 1 b = 4 c = -1 so x = 8 +- i
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
y = 4x-3
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
No.
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
Assuming that by writing that equation as a question, you mean to ask how to solve it for x, that can be done as follows: x2 + 4x - 6 = 0 x2 + 4x + 4 = 10 (x + 2)2 = 10 x + 2 = √20 x = -2 ± √20
x^2+4x+7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
y=-x^2+4x-3