What are the solutions to the equation x^2-2x^2-4x+8=0 ?

The solutions to the equation x^2-2x^2-4x+8=0 are x=-2(1+sqrt(3)),x=2(sqrt(3)-1)

Find the zeros of x^2-2x^2-4x+8=0

The zeros of x^2-2x^2-4x+8=0 are x=-2(1+sqrt(3)),x=2(sqrt(3)-1)

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x^{2} - 2 x^{2} - 4 x + 8 = 0

x = - 2 (1 + 3), x = 2 (3 - 1)

Decimal Notation

x = - 5.46410 \dots, x = 1.46410 \dots

Solution steps

x^{2} - 2 x^{2} - 4 x + 8 = 0

Refine

- x^{2} - 4 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) \cdot 8}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) \cdot 8 = 43

x_{1, 2} = \frac{- ( - 4 ) \pm 4 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 3}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 4 3}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 4 3}{2 ( - 1 )} : - 2 (1 + 3)

x = \frac{- ( - 4 ) - 4 3}{2 ( - 1 )} : 2 (3 - 1)

The solutions to the quadratic equation are:

x = - 2 (1 + 3), x = 2 (3 - 1)

2 x^{2} + 6 x = 31

5 (2 x - 4) = - 4 (x^{2})

(x + 1) (x - 3) = 2

4 = 2 x^{2} + 3

4.6 y^{2} + 3.5 y - 1.75 = 0

What are the solutions to the equation x^2-2x^2-4x+8=0 ?
The solutions to the equation x^2-2x^2-4x+8=0 are x=-2(1+sqrt(3)),x=2(sqrt(3)-1)
Find the zeros of x^2-2x^2-4x+8=0
The zeros of x^2-2x^2-4x+8=0 are x=-2(1+sqrt(3)),x=2(sqrt(3)-1)