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

The solutions to the equation (5x+2)(x+1)=1 are x=(-7+sqrt(29))/(10),x=(-7-sqrt(29))/(10)

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

The zeros of (5x+2)(x+1)=1 are x=(-7+sqrt(29))/(10),x=(-7-sqrt(29))/(10)

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

x = \frac{- 7 + 29}{10}, x = \frac{- 7 - 29}{10}

Solution steps

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

Expand

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

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

Move

1

to the left side

5 x^{2} + 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

7^{2} - 4 \cdot 5 \cdot 1 = 29

x_{1, 2} = \frac{- 7 \pm 29}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 7 + 29}{2 \cdot 5}, x_{2} = \frac{- 7 - 29}{2 \cdot 5}

x = \frac{- 7 + 29}{2 \cdot 5} : \frac{- 7 + 29}{10}

x = \frac{- 7 - 29}{2 \cdot 5} : \frac{- 7 - 29}{10}

The solutions to the quadratic equation are:

x = \frac{- 7 + 29}{10}, x = \frac{- 7 - 29}{10}

3 (3)

(x - 1)^{2} = 25

- 5 \geq \frac{5 - 3 x}{2}

z^{2} + 8 z + 15

2 < 5 x - 8 \leq 12

What are the solutions to the equation (5x+2)(x+1)=1 ?
The solutions to the equation (5x+2)(x+1)=1 are x=(-7+sqrt(29))/(10),x=(-7-sqrt(29))/(10)
Find the zeros of (5x+2)(x+1)=1
The zeros of (5x+2)(x+1)=1 are x=(-7+sqrt(29))/(10),x=(-7-sqrt(29))/(10)