What are the solutions to the equation (2x+1)(x+3)=-1 ?

The solutions to the equation (2x+1)(x+3)=-1 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4

Find the zeros of (2x+1)(x+3)=-1

The zeros of (2x+1)(x+3)=-1 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4

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(2 x + 1) (x + 3) = - 1

x = \frac{- 7 + 17}{4}, x = \frac{- 7 - 17}{4}

Decimal Notation

x = - 0.71922 \dots, x = - 2.78077 \dots

Solution steps

(2 x + 1) (x + 3) = - 1

Expand

(2 x + 1) (x + 3) : 2 x^{2} + 7 x + 3

2 x^{2} + 7 x + 3 = - 1

Move

1

to the left side

2 x^{2} + 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 4 = 17

x_{1, 2} = \frac{- 7 \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 17}{2 \cdot 2}, x_{2} = \frac{- 7 - 17}{2 \cdot 2}

x = \frac{- 7 + 17}{2 \cdot 2} : \frac{- 7 + 17}{4}

x = \frac{- 7 - 17}{2 \cdot 2} : \frac{- 7 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 17}{4}, x = \frac{- 7 - 17}{4}

x (x - 2) > 0

\frac{1}{8} + \frac{1}{2}

r^{2} + 2 r + 14 = 0

2 x + 5 \leq 15

ln (e) = x

What are the solutions to the equation (2x+1)(x+3)=-1 ?
The solutions to the equation (2x+1)(x+3)=-1 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4
Find the zeros of (2x+1)(x+3)=-1
The zeros of (2x+1)(x+3)=-1 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4