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

The solutions to the equation (5x+1)(x+2)=-1 are x=(-11+sqrt(61))/(10),x=(-11-sqrt(61))/(10)

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

The zeros of (5x+1)(x+2)=-1 are x=(-11+sqrt(61))/(10),x=(-11-sqrt(61))/(10)

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

x = \frac{- 11 + 61}{10}, x = \frac{- 11 - 61}{10}

Decimal Notation

x = - 0.31897 \dots, x = - 1.88102 \dots

Solution steps

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

Expand

(5 x + 1) (x + 2) : 5 x^{2} + 11 x + 2

5 x^{2} + 11 x + 2 = - 1

Move

1

to the left side

5 x^{2} + 11 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

1 1^{2} - 4 \cdot 5 \cdot 3 = 61

x_{1, 2} = \frac{- 11 \pm 61}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 11 + 61}{2 \cdot 5}, x_{2} = \frac{- 11 - 61}{2 \cdot 5}

x = \frac{- 11 + 61}{2 \cdot 5} : \frac{- 11 + 61}{10}

x = \frac{- 11 - 61}{2 \cdot 5} : \frac{- 11 - 61}{10}

The solutions to the quadratic equation are:

x = \frac{- 11 + 61}{10}, x = \frac{- 11 - 61}{10}

x^{2} - 10 x + 61 = 0

3 x^{2} + 6 x + 4 = 0

4 x + 3 > 15 or - 6 x \geq 12

\frac{2 x ^{- 2} y}{7 x ^{3} y ^{2}} \cdot \frac{14 x ^{5} y ^{3}}{x ^{0} y ^{5}}

x^{2} - 4 x - 1

What are the solutions to the equation (5x+1)(x+2)=-1 ?
The solutions to the equation (5x+1)(x+2)=-1 are x=(-11+sqrt(61))/(10),x=(-11-sqrt(61))/(10)
Find the zeros of (5x+1)(x+2)=-1
The zeros of (5x+1)(x+2)=-1 are x=(-11+sqrt(61))/(10),x=(-11-sqrt(61))/(10)