What are the solutions to the equation (5x+4)(x+1)=2 ?

The solutions to the equation (5x+4)(x+1)=2 are x=(-9+sqrt(41))/(10),x=(-9-sqrt(41))/(10)

Find the zeros of (5x+4)(x+1)=2

The zeros of (5x+4)(x+1)=2 are x=(-9+sqrt(41))/(10),x=(-9-sqrt(41))/(10)

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(5 x + 4) (x + 1) = 2

x = \frac{- 9 + 41}{10}, x = \frac{- 9 - 41}{10}

Decimal Notation

x = - 0.25968 \dots, x = - 1.54031 \dots

Solution steps

(5 x + 4) (x + 1) = 2

Expand

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

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

Move

2

to the left side

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

Solve with the quadratic formula

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

9^{2} - 4 \cdot 5 \cdot 2 = 41

x_{1, 2} = \frac{- 9 \pm 41}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 9 + 41}{2 \cdot 5}, x_{2} = \frac{- 9 - 41}{2 \cdot 5}

x = \frac{- 9 + 41}{2 \cdot 5} : \frac{- 9 + 41}{10}

x = \frac{- 9 - 41}{2 \cdot 5} : \frac{- 9 - 41}{10}

The solutions to the quadratic equation are:

x = \frac{- 9 + 41}{10}, x = \frac{- 9 - 41}{10}

3 - 2 x^{2} = 7

4 (x^{2} + 2 x - 4) = 0

q (q + 7) = 5 q^{2} - 3

3 x^{2} - x^{2} + 7 x + 3 = 4

3 x^{2} - 6 x - 22 = 0

What are the solutions to the equation (5x+4)(x+1)=2 ?
The solutions to the equation (5x+4)(x+1)=2 are x=(-9+sqrt(41))/(10),x=(-9-sqrt(41))/(10)
Find the zeros of (5x+4)(x+1)=2
The zeros of (5x+4)(x+1)=2 are x=(-9+sqrt(41))/(10),x=(-9-sqrt(41))/(10)