What are the solutions to the equation (5-x)^2+2x^2+8x-20=0 ?

The solutions to the equation (5-x)^2+2x^2+8x-20=0 are x= 1/3+i(sqrt(14))/3 ,x= 1/3-i(sqrt(14))/3

Find the zeros of (5-x)^2+2x^2+8x-20=0

The zeros of (5-x)^2+2x^2+8x-20=0 are x= 1/3+i(sqrt(14))/3 ,x= 1/3-i(sqrt(14))/3

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(5 - x)^{2} + 2 x^{2} + 8 x - 20 = 0

x = \frac{1}{3} + i \frac{14}{3}, x = \frac{1}{3} - i \frac{14}{3}

Solution steps

(5 - x)^{2} + 2 x^{2} + 8 x - 20 = 0

Expand

(5 - x)^{2} + 2 x^{2} + 8 x - 20 : 3 x^{2} - 2 x + 5

3 x^{2} - 2 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

Simplify

(- 2)^{2} - 4 \cdot 3 \cdot 5 : 214 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 14 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 14 i}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 14 i}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 14 i}{2 \cdot 3} : \frac{1}{3} + i \frac{14}{3}

x = \frac{- ( - 2 ) - 2 14 i}{2 \cdot 3} : \frac{1}{3} - i \frac{14}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{14}{3}, x = \frac{1}{3} - i \frac{14}{3}

x^{2} - 2 x - 323 = 0

2 x^{2} - 8 x - 16

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

2 y^{2} - 8 = 6 - 2 y^{2}

x, \frac{x ^{2}}{18} = R

What are the solutions to the equation (5-x)^2+2x^2+8x-20=0 ?
The solutions to the equation (5-x)^2+2x^2+8x-20=0 are x= 1/3+i(sqrt(14))/3 ,x= 1/3-i(sqrt(14))/3
Find the zeros of (5-x)^2+2x^2+8x-20=0
The zeros of (5-x)^2+2x^2+8x-20=0 are x= 1/3+i(sqrt(14))/3 ,x= 1/3-i(sqrt(14))/3