What are the solutions to the equation (x+3)(2x+1)=9 ?

The solutions to the equation (x+3)(2x+1)=9 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

Find the zeros of (x+3)(2x+1)=9

The zeros of (x+3)(2x+1)=9 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

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(x + 3) (2 x + 1) = 9

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

Solution steps

(x + 3) (2 x + 1) = 9

Expand

(x + 3) (2 x + 1) : 2 x^{2} + 7 x + 3

2 x^{2} + 7 x + 3 = 9

Move

9

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 ( - 6 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 6) = 97

x_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 97}{2 \cdot 2}, x_{2} = \frac{- 7 - 97}{2 \cdot 2}

x = \frac{- 7 + 97}{2 \cdot 2} : \frac{- 7 + 97}{4}

x = \frac{- 7 - 97}{2 \cdot 2} : \frac{- 7 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

- 2 + 3 (- 4) \geq \frac{1}{3}

25 - 9 w^{2}

12 x = 48

6 x^{2} + 2 x = 0

What are the solutions to the equation (x+3)(2x+1)=9 ?
The solutions to the equation (x+3)(2x+1)=9 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4
Find the zeros of (x+3)(2x+1)=9
The zeros of (x+3)(2x+1)=9 are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4