What are the solutions to the equation (x^2+2x^2+3)=(x+3x+x^2-5) ?

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

Find the zeros of (x^2+2x^2+3)=(x+3x+x^2-5)

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

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

x = 1 - 3 i, x = 1 + 3 i

Solution steps

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

Refine

3 x^{2} + 3 = 4 x + x^{2} - 5

Switch sides

4 x + x^{2} - 5 = 3 x^{2} + 3

Move

3

to the left side

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

Move

3 x^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

4^{2} - 4 (- 2) (- 8) : 43 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - 3 i, x = 1 + 3 i

y, x^{2} + y^{2} = a^{2}

2 x^{2} - 144 = - 2 x

x^{2} + 1.4 x + 1 = 0

10000 x^{2} - 64 = 0

x, x^{2} - 8 x + 41 = 0

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