What are the solutions to the equation (2x+3)(x+4)=67 ?

The solutions to the equation (2x+3)(x+4)=67 are x=(-11+sqrt(561))/4 ,x=(-11-sqrt(561))/4

Find the zeros of (2x+3)(x+4)=67

The zeros of (2x+3)(x+4)=67 are x=(-11+sqrt(561))/4 ,x=(-11-sqrt(561))/4

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(2 x + 3) (x + 4) = 67

x = \frac{- 11 + 561}{4}, x = \frac{- 11 - 561}{4}

Decimal Notation

x = 3.17135 \dots, x = - 8.67135 \dots

Solution steps

(2 x + 3) (x + 4) = 67

Expand

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

2 x^{2} + 11 x + 12 = 67

Move

67

to the left side

2 x^{2} + 11 x - 55 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 ( - 55 )}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 (- 55) = 561

x_{1, 2} = \frac{- 11 \pm 561}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 561}{2 \cdot 2}, x_{2} = \frac{- 11 - 561}{2 \cdot 2}

x = \frac{- 11 + 561}{2 \cdot 2} : \frac{- 11 + 561}{4}

x = \frac{- 11 - 561}{2 \cdot 2} : \frac{- 11 - 561}{4}

The solutions to the quadratic equation are:

x = \frac{- 11 + 561}{4}, x = \frac{- 11 - 561}{4}

v^{2} = - v + 30

4 m^{2} - 6 m - 14 = 10

x, f = x^{2} - 8 x + m

5 x^{2} - 8 x - 1 = 0

x^{2} - 160 x + 3200 = 0

What are the solutions to the equation (2x+3)(x+4)=67 ?
The solutions to the equation (2x+3)(x+4)=67 are x=(-11+sqrt(561))/4 ,x=(-11-sqrt(561))/4
Find the zeros of (2x+3)(x+4)=67
The zeros of (2x+3)(x+4)=67 are x=(-11+sqrt(561))/4 ,x=(-11-sqrt(561))/4