What are the solutions to the equation (5x+4)(x+1)=1 ?

The solutions to the equation (5x+4)(x+1)=1 are x=(-9+sqrt(21))/(10),x=(-9-sqrt(21))/(10)

Find the zeros of (5x+4)(x+1)=1

The zeros of (5x+4)(x+1)=1 are x=(-9+sqrt(21))/(10),x=(-9-sqrt(21))/(10)

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(5 x + 4) (x + 1) = 1

x = \frac{- 9 + 21}{10}, x = \frac{- 9 - 21}{10}

Decimal Notation

x = - 0.44174 \dots, x = - 1.35825 \dots

Solution steps

(5 x + 4) (x + 1) = 1

Expand

(5 x + 4) (x + 1) : 5 x^{2} + 9 x + 4

5 x^{2} + 9 x + 4 = 1

Move

1

to the left side

5 x^{2} + 9 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

9^{2} - 4 \cdot 5 \cdot 3 = 21

x_{1, 2} = \frac{- 9 \pm 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 9 + 21}{2 \cdot 5}, x_{2} = \frac{- 9 - 21}{2 \cdot 5}

x = \frac{- 9 + 21}{2 \cdot 5} : \frac{- 9 + 21}{10}

x = \frac{- 9 - 21}{2 \cdot 5} : \frac{- 9 - 21}{10}

The solutions to the quadratic equation are:

x = \frac{- 9 + 21}{10}, x = \frac{- 9 - 21}{10}

x^{\frac{1}{3}} x^{\frac{1}{3}}

12 x^{2} = 300

- x^{2} + 120 = 0

7 m + 11 - 5 m - 2 = 2 m - 9

11 \geq y + 10

What are the solutions to the equation (5x+4)(x+1)=1 ?
The solutions to the equation (5x+4)(x+1)=1 are x=(-9+sqrt(21))/(10),x=(-9-sqrt(21))/(10)
Find the zeros of (5x+4)(x+1)=1
The zeros of (5x+4)(x+1)=1 are x=(-9+sqrt(21))/(10),x=(-9-sqrt(21))/(10)