What are the solutions to the equation 2(x^2-2x)^2-5(x^2-2x)=7 ?

The solutions to the equation 2(x^2-2x)^2-5(x^2-2x)=7 are x=1,x=(2+3sqrt(2))/2 ,x=(2-3sqrt(2))/2

Find the zeros of 2(x^2-2x)^2-5(x^2-2x)=7

The zeros of 2(x^2-2x)^2-5(x^2-2x)=7 are x=1,x=(2+3sqrt(2))/2 ,x=(2-3sqrt(2))/2

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2 (x^{2} - 2 x)^{2} - 5 (x^{2} - 2 x) = 7

x = 1, x = \frac{2 + 3 2}{2}, x = \frac{2 - 3 2}{2}

Decimal Notation

x = 1, x = 3.12132 \dots, x = - 1.12132 \dots

Solution steps

2 (x^{2} - 2 x)^{2} - 5 (x^{2} - 2 x) = 7

Expand

2 (x^{2} - 2 x)^{2} - 5 (x^{2} - 2 x) : 2 x^{4} - 8 x^{3} + 3 x^{2} + 10 x

2 x^{4} - 8 x^{3} + 3 x^{2} + 10 x = 7

Move

7

to the left side

2 x^{4} - 8 x^{3} + 3 x^{2} + 10 x - 7 = 0

Factor

2 x^{4} - 8 x^{3} + 3 x^{2} + 10 x - 7 : (x - 1)^{2} (2 x^{2} - 4 x - 7)

(x - 1)^{2} (2 x^{2} - 4 x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 2 x^{2} - 4 x - 7 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x^{2} - 4 x - 7 = 0 : x = \frac{2 + 3 2}{2}, x = \frac{2 - 3 2}{2}

The solutions are

x = 1, x = \frac{2 + 3 2}{2}, x = \frac{2 - 3 2}{2}

z^{6} + z + 1 = 0

(x - 3) (x + 1)^{2} (x^{2} - 4 x + 13) = 0

4 y^{3} - 2 y + 5 = 0

6 x^{4} - 81 x^{2} + 162 = 0

x, f = x^{3} h

What are the solutions to the equation 2(x^2-2x)^2-5(x^2-2x)=7 ?
The solutions to the equation 2(x^2-2x)^2-5(x^2-2x)=7 are x=1,x=(2+3sqrt(2))/2 ,x=(2-3sqrt(2))/2
Find the zeros of 2(x^2-2x)^2-5(x^2-2x)=7
The zeros of 2(x^2-2x)^2-5(x^2-2x)=7 are x=1,x=(2+3sqrt(2))/2 ,x=(2-3sqrt(2))/2