What are the solutions to the equation (2x+3)(x+4)=1 ?

The solutions to the equation (2x+3)(x+4)=1 are x=(-11+sqrt(33))/4 ,x=(-11-sqrt(33))/4

Find the zeros of (2x+3)(x+4)=1

The zeros of (2x+3)(x+4)=1 are x=(-11+sqrt(33))/4 ,x=(-11-sqrt(33))/4

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(2 x + 3) (x + 4) = 1

x = \frac{- 11 + 33}{4}, x = \frac{- 11 - 33}{4}

Decimal Notation

x = - 1.31385 \dots, x = - 4.18614 \dots

Solution steps

(2 x + 3) (x + 4) = 1

Expand

(2 x + 3) (x + 4) : 2 x^{2} + 11 x + 12

2 x^{2} + 11 x + 12 = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

1 1^{2} - 4 \cdot 2 \cdot 11 = 33

x_{1, 2} = \frac{- 11 \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 33}{2 \cdot 2}, x_{2} = \frac{- 11 - 33}{2 \cdot 2}

x = \frac{- 11 + 33}{2 \cdot 2} : \frac{- 11 + 33}{4}

x = \frac{- 11 - 33}{2 \cdot 2} : \frac{- 11 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{- 11 + 33}{4}, x = \frac{- 11 - 33}{4}

\frac{x - 1}{x + 1} \leq \frac{x + 3}{x - 5}

x^{2} - 5 x = 14

(\frac{3 x ^{2} y ^{5} z ^{- 2}}{5 x z ^{5}})^{- 3}

5 x^{2} + 6 x = - 5

y, 2 x - y + 5^{x - 1} = 7 \cdot 5

What are the solutions to the equation (2x+3)(x+4)=1 ?
The solutions to the equation (2x+3)(x+4)=1 are x=(-11+sqrt(33))/4 ,x=(-11-sqrt(33))/4
Find the zeros of (2x+3)(x+4)=1
The zeros of (2x+3)(x+4)=1 are x=(-11+sqrt(33))/4 ,x=(-11-sqrt(33))/4