x2 + 12x - 64 = 0 ∴ (x + 16)(x - 4) = 0 ∴ x ∈ {4, -16}
x2 - 10x + 25 = 64 Or x2 - 10x + 25 = 64? If the first, move all over to 1 side; x2 - 10x + 25 - 64 = 0 x2 - 10x + -39 = 0 Then make it into quadratic equations; (x - 13)(x + 3) = 0 Therefore x = 13 or x = -3. If the second, 8x = -39 -39/8 = -9.75 = x
144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x: x2 - 81 = 0
x2 + 12x - 64 = 0 ∴ (x + 16)(x - 4) = 0 ∴ x ∈ {4, -16}
x2 - 10x + 25 = 64 Or x2 - 10x + 25 = 64? If the first, move all over to 1 side; x2 - 10x + 25 - 64 = 0 x2 - 10x + -39 = 0 Then make it into quadratic equations; (x - 13)(x + 3) = 0 Therefore x = 13 or x = -3. If the second, 8x = -39 -39/8 = -9.75 = x
144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
x2-2x-3 = 0 (x+1)(x-3) = 0 x = -1 or x = 3
Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + 7x = 0 => x*(x + 7) = 0 => x = 0 or x + 7 = 0 so that x = 0 or x = -7
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)
x2-4x-21 = 0 => x = -3 or x = 7 x2-3x-18 = 0 => x = -3 or x = 6