What are the solutions to the equation 2(x+5)^2+(x-3)^2=14(x+4) ?

The solutions to the equation 2(x+5)^2+(x-3)^2=14(x+4) are x=i,x=-i

Find the zeros of 2(x+5)^2+(x-3)^2=14(x+4)

The zeros of 2(x+5)^2+(x-3)^2=14(x+4) are x=i,x=-i

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2 (x + 5)^{2} + (x - 3)^{2} = 14 (x + 4)

x = i, x = - i

Solution steps

2 (x + 5)^{2} + (x - 3)^{2} = 14 (x + 4)

Expand

2 (x + 5)^{2} + (x - 3)^{2} : 3 x^{2} + 14 x + 59

Expand

14 (x + 4) : 14 x + 56

3 x^{2} + 14 x + 59 = 14 x + 56

Move

59

to the right side

3 x^{2} + 14 x = 14 x - 3

Move

14 x

to the left side

3 x^{2} = - 3

Divide both sides by

3

x^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - 1, x = - - 1

- 1 = i

- - 1 = - i

x = i, x = - i

96 t - 16 t^{2} = 0

y^{2} - 5 y - 6

\frac{1}{e ^{- 3}}

x^{2} + x y - 12 y^{2}

- 3 - 12

What are the solutions to the equation 2(x+5)^2+(x-3)^2=14(x+4) ?
The solutions to the equation 2(x+5)^2+(x-3)^2=14(x+4) are x=i,x=-i
Find the zeros of 2(x+5)^2+(x-3)^2=14(x+4)
The zeros of 2(x+5)^2+(x-3)^2=14(x+4) are x=i,x=-i