What are the solutions to the equation (x+3)2=8(x-2)+x^2 ?

The solutions to the equation (x+3)2=8(x-2)+x^2 are x=-3+sqrt(31),x=-3-sqrt(31)

Find the zeros of (x+3)2=8(x-2)+x^2

The zeros of (x+3)2=8(x-2)+x^2 are x=-3+sqrt(31),x=-3-sqrt(31)

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(x + 3) 2 = 8 (x - 2) + x^{2}

x = - 3 + 31, x = - 3 - 31

Solution steps

(x + 3) \cdot 2 = 8 (x - 2) + x^{2}

Expand

(x + 3) \cdot 2 : 2 x + 6

Expand

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

2 x + 6 = 8 x - 16 + x^{2}

Switch sides

8 x - 16 + x^{2} = 2 x + 6

Move

6

to the left side

x^{2} + 8 x - 22 = 2 x

Move

2 x

to the left side

x^{2} + 6 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 22 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 22) = 231

x_{1, 2} = \frac{- 6 \pm 2 31}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 2 31}{2 \cdot 1}, x_{2} = \frac{- 6 - 2 31}{2 \cdot 1}

x = \frac{- 6 + 2 31}{2 \cdot 1} : - 3 + 31

x = \frac{- 6 - 2 31}{2 \cdot 1} : - 3 - 31

The solutions to the quadratic equation are:

x = - 3 + 31, x = - 3 - 31

x^{2} + 13 x + 24 = 0

2 b^{2} + 40 = 0

(\frac{5}{13})^{2} + x^{2} = 1

6 s^{2} + 18 s = 0

5 x^{2} - 72 = 4 x^{2}

What are the solutions to the equation (x+3)2=8(x-2)+x^2 ?
The solutions to the equation (x+3)2=8(x-2)+x^2 are x=-3+sqrt(31),x=-3-sqrt(31)
Find the zeros of (x+3)2=8(x-2)+x^2
The zeros of (x+3)2=8(x-2)+x^2 are x=-3+sqrt(31),x=-3-sqrt(31)