What are the solutions to the equation 2a-4x=x^2+a^2-4 ?

The solutions to the equation 2a-4x=x^2+a^2-4 are x=-2+sqrt(-a^2+2a+8),x=-sqrt(-a^2+2a+8)-2

Find the zeros of 2a-4x=x^2+a^2-4

The zeros of 2a-4x=x^2+a^2-4 are x=-2+sqrt(-a^2+2a+8),x=-sqrt(-a^2+2a+8)-2

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2 a - 4 x = x^{2} + a^{2} - 4

x = - 2 + - a^{2} + 2 a + 8, x = - - a^{2} + 2 a + 8 - 2

Solution steps

2 a - 4 x = x^{2} + a^{2} - 4

Switch sides

x^{2} + a^{2} - 4 = 2 a - 4 x

Move

4 x

to the left side

x^{2} + a^{2} - 4 + 4 x = 2 a

Move

2 a

to the left side

x^{2} + a^{2} - 4 + 4 x - 2 a = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + 4 x + a^{2} - 4 - 2 a = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( a ^{2} - 4 - 2 a )}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot (a^{2} - 4 - 2 a) : 2 - a^{2} + 2 a + 8

x_{1, 2} = \frac{- 4 \pm 2 - a ^{2} + 2 a + 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 - a ^{2} + 2 a + 8}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 - a ^{2} + 2 a + 8}{2 \cdot 1}

x = \frac{- 4 + 2 - a ^{2} + 2 a + 8}{2 \cdot 1} : - 2 + - a^{2} + 2 a + 8

x = \frac{- 4 - 2 - a ^{2} + 2 a + 8}{2 \cdot 1} : - - a^{2} + 2 a + 8 - 2

The solutions to the quadratic equation are:

x = - 2 + - a^{2} + 2 a + 8, x = - - a^{2} + 2 a + 8 - 2

0.0493 x^{2} - 0.9621 x + 53.104 - 70 = 0

(2 x + 1)^{2} = 1 + (x + 1) (x - 1)

x^{2} - 2 x + 6

4 x^{2} - 13 = 9 x

(x + 5)^{2} - 3 (x + 5) - 10 = 0

What are the solutions to the equation 2a-4x=x^2+a^2-4 ?
The solutions to the equation 2a-4x=x^2+a^2-4 are x=-2+sqrt(-a^2+2a+8),x=-sqrt(-a^2+2a+8)-2
Find the zeros of 2a-4x=x^2+a^2-4
The zeros of 2a-4x=x^2+a^2-4 are x=-2+sqrt(-a^2+2a+8),x=-sqrt(-a^2+2a+8)-2