What are the solutions to the equation (x+4)^2=15-2x^2 ?

The solutions to the equation (x+4)^2=15-2x^2 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3

Find the zeros of (x+4)^2=15-2x^2

The zeros of (x+4)^2=15-2x^2 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3

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(x + 4)^{2} = 15 - 2 x^{2}

x = \frac{- 4 + 13}{3}, x = - \frac{4 + 13}{3}

Decimal Notation

x = - 0.13148 \dots, x = - 2.53518 \dots

Solution steps

(x + 4)^{2} = 15 - 2 x^{2}

Expand

(x + 4)^{2} : x^{2} + 8 x + 16

x^{2} + 8 x + 16 = 15 - 2 x^{2}

Move

2 x^{2}

to the left side

3 x^{2} + 8 x + 16 = 15

Move

15

to the left side

3 x^{2} + 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

8^{2} - 4 \cdot 3 \cdot 1 = 213

x_{1, 2} = \frac{- 8 \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 8 + 2 13}{2 \cdot 3}, x_{2} = \frac{- 8 - 2 13}{2 \cdot 3}

x = \frac{- 8 + 2 13}{2 \cdot 3} : \frac{- 4 + 13}{3}

x = \frac{- 8 - 2 13}{2 \cdot 3} : - \frac{4 + 13}{3}

The solutions to the quadratic equation are:

x = \frac{- 4 + 13}{3}, x = - \frac{4 + 13}{3}

r^{2} + 16 r + 64 = 0

p^{2} - 49 = 0

(2 x + c)^{2} = 2 x - c

x, 9 x^{2} - 6 x y + y^{2} - 80 x - 40 y = 0

7 x^{2} - 2 x - 9 = 0

What are the solutions to the equation (x+4)^2=15-2x^2 ?
The solutions to the equation (x+4)^2=15-2x^2 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3
Find the zeros of (x+4)^2=15-2x^2
The zeros of (x+4)^2=15-2x^2 are x=(-4+sqrt(13))/3 ,x=-(4+sqrt(13))/3