What are the solutions to the equation (2x+5)(x+1)=-1 ?

The solutions to the equation (2x+5)(x+1)=-1 are x=-3/2 ,x=-2

Find the zeros of (2x+5)(x+1)=-1

The zeros of (2x+5)(x+1)=-1 are x=-3/2 ,x=-2

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(2 x + 5) (x + 1) = - 1

x = - \frac{3}{2}, x = - 2

Decimal Notation

x = - 1.5, x = - 2

Solution steps

(2 x + 5) (x + 1) = - 1

Expand

(2 x + 5) (x + 1) : 2 x^{2} + 7 x + 5

2 x^{2} + 7 x + 5 = - 1

Move

1

to the left side

2 x^{2} + 7 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 6 = 1

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

Separate the solutions

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

x = \frac{- 7 + 1}{2 \cdot 2} : - \frac{3}{2}

x = \frac{- 7 - 1}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = - 2

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

20 = 0.36 x + 15.9

y^{2} - 8 y + 16 = 0

3 - x > 0

\frac{4}{5 ^{- 2}}

What are the solutions to the equation (2x+5)(x+1)=-1 ?
The solutions to the equation (2x+5)(x+1)=-1 are x=-3/2 ,x=-2
Find the zeros of (2x+5)(x+1)=-1
The zeros of (2x+5)(x+1)=-1 are x=-3/2 ,x=-2