If you actually meant that as you wrote it:
x2 = 7 - 4
Then the answer is simple:
x2 = 7 - 4
∴ x2 = 3
∴ x = √3
If on the other hand, you actually wanted to say:
x2 = 7x - 4
Then you can find it this way:
x2 = 7x - 4
∴ x - 7x = -4
∴ x - 7x + (7/2)2 = -4 + (7/2)2
∴ (x - 7/2)2 = -16/4 + 49/4
∴ x - 7/2 = ± √(33/4)
∴ x = 7/2 ± √33/2
∴ x = (7 ± √33) / 2
Alternatively, if you meant to say:
x2 = 7 - 4x
Then the answer is:
x2 = 7 - 4x
∴ x2 + 4x = 7
∴ x2 + 4x + 4 = 7
∴ (x + 2)2 = 7
∴ x + 2 = ± √7
∴ x = -2 ± √7
x=18
(2x)ysquared
144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144
x = +13x = - 13
You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.
x2 = 6482 = 64x = 8
x: x2 - 81 = 0
x=18
(2x)ysquared
Divide both sides of the equation by x: x2 = 9x x2 / x = 9x / x x = 9
144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144
x = +13x = - 13
You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.
X2 = k Take square root each side. X = (+/-) sqrt(k) ============
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)