Top Answer

No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola,

and every point on the parabola satisfies the equation.

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0x2+2x-63 = 0 When factorised: (x-7)(x+9) = 0 Therefore: x = 7 or x = -9

x2+2x-63 = 0 (x-7)(x+9) = 0 x = 7 or x = -9 Use the quadratic equation formula

x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.

x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}

(x - 7)(x + 9) = x2 + 2x - 63

This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========

(4 ± i2) where i2 = -1

x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}

x2+2x-63 = (x-7)(x+9) when factored

x2 = 6482 = 64x = 8

Equals anything... x is a variable. If that equation was set equal to zero then you could solve for x, but that is not what you have asked.

the question is to solve (x^2-5x+5)^(x^2-36)=1

Simplifying x2 + -2x + -63 Reorder the terms: -63 + -2x + x2 Factor a trinomial. (-7 + -1x)(9 + -1x) Final result: (-7 + -1x)(9 + -1x)

x2 + 6x + 8 = 0 Solve for x.X = -2 or X = -4

2x+6x=-9 => 8x=-9=> x=-8/9

There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).

x2-13x+36=(x-9)(x-4)=0 x=9 or x=4

If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2

x(x - 1) = x2 + x x2 - x = x2 + x The only solution to this equation is x = 0

The quadratic expression x2+6x+8 when factorised equals (x+2)(x+4)

X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2

A quadratic equation. If you wish to solve for x, you can do so as follows: -x2 + 6x + 7 = 0 x2 - 6x - 7 = 0 (x - 7)(x + 1) = 0 x ∈ {-1, 7}

x2 + 3y = 7 3x + y2 = 3 3y = x2 + 7 y2 = -3x + 3 y = x2/3 + 7/3 y = ± √(-3x + 3) If you draw the graphs of y = x2/3 + 7/3 and y = ± √(-3x + 3) in a graphing calculator, you will see that they don't intersect, so that the system of the given equations has not a solution.

x2 + 36x = 0 x2 = -36x x = -36

if x2 + 7 = 37, then x2 = 29 and x = ±√29

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