What are the solutions to the equation (2x+4)(x+7)=2 ?

The solutions to the equation (2x+4)(x+7)=2 are x=(-9+sqrt(29))/2 ,x=-(9+sqrt(29))/2

Find the zeros of (2x+4)(x+7)=2

The zeros of (2x+4)(x+7)=2 are x=(-9+sqrt(29))/2 ,x=-(9+sqrt(29))/2

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(2 x + 4) (x + 7) = 2

x = \frac{- 9 + 29}{2}, x = - \frac{9 + 29}{2}

Decimal Notation

x = - 1.80741 \dots, x = - 7.19258 \dots

Solution steps

(2 x + 4) (x + 7) = 2

Expand

(2 x + 4) (x + 7) : 2 x^{2} + 18 x + 28

2 x^{2} + 18 x + 28 = 2

Move

2

to the left side

2 x^{2} + 18 x + 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 2 \cdot 26}{2 \cdot 2}

1 8^{2} - 4 \cdot 2 \cdot 26 = 229

x_{1, 2} = \frac{- 18 \pm 2 29}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 18 + 2 29}{2 \cdot 2}, x_{2} = \frac{- 18 - 2 29}{2 \cdot 2}

x = \frac{- 18 + 2 29}{2 \cdot 2} : \frac{- 9 + 29}{2}

x = \frac{- 18 - 2 29}{2 \cdot 2} : - \frac{9 + 29}{2}

The solutions to the quadratic equation are:

x = \frac{- 9 + 29}{2}, x = - \frac{9 + 29}{2}

- x^{2} + 6 = 0

- 16 x^{2} = 1

x^{2} + 7 x + 2

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

x^{2} + 65 = 209

What are the solutions to the equation (2x+4)(x+7)=2 ?
The solutions to the equation (2x+4)(x+7)=2 are x=(-9+sqrt(29))/2 ,x=-(9+sqrt(29))/2
Find the zeros of (2x+4)(x+7)=2
The zeros of (2x+4)(x+7)=2 are x=(-9+sqrt(29))/2 ,x=-(9+sqrt(29))/2