What are the solutions to the equation 2x^2-(4x^2-4x-4)=1160 ?

The solutions to the equation 2x^2-(4x^2-4x-4)=1160 are x=1-sqrt(577)i,x=1+sqrt(577)i

Find the zeros of 2x^2-(4x^2-4x-4)=1160

The zeros of 2x^2-(4x^2-4x-4)=1160 are x=1-sqrt(577)i,x=1+sqrt(577)i

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2 x^{2} - (4 x^{2} - 4 x - 4) = 1160

x = 1 - 577 i, x = 1 + 577 i

Solution steps

2 x^{2} - (4 x^{2} - 4 x - 4) = 1160

Expand

2 x^{2} - (4 x^{2} - 4 x - 4) : - 2 x^{2} + 4 x + 4

- 2 x^{2} + 4 x + 4 = 1160

Move

1160

to the left side

- 2 x^{2} + 4 x - 1156 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 2 ) ( - 1156 )}{2 ( - 2 )}

Simplify

4^{2} - 4 (- 2) (- 1156) : 4577 i

x_{1, 2} = \frac{- 4 \pm 4 577 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 4 + 4 577 i}{2 ( - 2 )}, x_{2} = \frac{- 4 - 4 577 i}{2 ( - 2 )}

x = \frac{- 4 + 4 577 i}{2 ( - 2 )} : 1 - 577 i

x = \frac{- 4 - 4 577 i}{2 ( - 2 )} : 1 + 577 i

The solutions to the quadratic equation are:

x = 1 - 577 i, x = 1 + 577 i

(x + 2)^{2} + x^{2} = 5 8^{2}

- 10 (- 3 x^{2} - 1) = 0

x^{2} - 3 (a + 2) x + b = 2 x - 3 a

x^{2} + 27 x + 50 = 0

4 x^{2} + 8 = 28

What are the solutions to the equation 2x^2-(4x^2-4x-4)=1160 ?
The solutions to the equation 2x^2-(4x^2-4x-4)=1160 are x=1-sqrt(577)i,x=1+sqrt(577)i
Find the zeros of 2x^2-(4x^2-4x-4)=1160
The zeros of 2x^2-(4x^2-4x-4)=1160 are x=1-sqrt(577)i,x=1+sqrt(577)i