What are the solutions to the equation 2(x^2+4x+4)+1=51 ?

The solutions to the equation 2(x^2+4x+4)+1=51 are x=3,x=-7

Find the zeros of 2(x^2+4x+4)+1=51

The zeros of 2(x^2+4x+4)+1=51 are x=3,x=-7

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2 (x^{2} + 4 x + 4) + 1 = 51

x = 3, x = - 7

Solution steps

2 (x^{2} + 4 x + 4) + 1 = 51

Expand

2 (x^{2} + 4 x + 4) + 1 : 2 x^{2} + 8 x + 9

2 x^{2} + 8 x + 9 = 51

Move

51

to the left side

2 x^{2} + 8 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 42) = 20

x_{1, 2} = \frac{- 8 \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 20}{2 \cdot 2}, x_{2} = \frac{- 8 - 20}{2 \cdot 2}

x = \frac{- 8 + 20}{2 \cdot 2} : 3

x = \frac{- 8 - 20}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

x = 3, x = - 7

- 6 x^{2} + 12 x - 6 = 0

n (n + 14) = - 4

4 x^{2} - 28 x = - 49

- x^{2} - 4 x + 32 = 0

12 a^{2} - 7 a = 12

What are the solutions to the equation 2(x^2+4x+4)+1=51 ?
The solutions to the equation 2(x^2+4x+4)+1=51 are x=3,x=-7
Find the zeros of 2(x^2+4x+4)+1=51
The zeros of 2(x^2+4x+4)+1=51 are x=3,x=-7