What are the solutions to the equation (x+3)(x+4)=6x ?

The solutions to the equation (x+3)(x+4)=6x are x=-1/2+i(sqrt(47))/2 ,x=-1/2-i(sqrt(47))/2

Find the zeros of (x+3)(x+4)=6x

The zeros of (x+3)(x+4)=6x are x=-1/2+i(sqrt(47))/2 ,x=-1/2-i(sqrt(47))/2

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(x + 3) (x + 4) = 6 x

x = - \frac{1}{2} + i \frac{47}{2}, x = - \frac{1}{2} - i \frac{47}{2}

Solution steps

(x + 3) (x + 4) = 6 x

Expand

(x + 3) (x + 4) : x^{2} + 7 x + 12

x^{2} + 7 x + 12 = 6 x

Move

6 x

to the left side

x^{2} + x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot 12 : 47 i

x_{1, 2} = \frac{- 1 \pm 47 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 47 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 47 i}{2 \cdot 1}

x = \frac{- 1 + 47 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{47}{2}

x = \frac{- 1 - 47 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{47}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{47}{2}, x = - \frac{1}{2} - i \frac{47}{2}

12 x - x^{2} + 3 = 9

x = 60 x - 0.01 \cdot x^{2}

4 x^{2} - 9 x = 71

1 x^{2} - 2 x = 0

4 x^{2} + 25 = 16 x

What are the solutions to the equation (x+3)(x+4)=6x ?
The solutions to the equation (x+3)(x+4)=6x are x=-1/2+i(sqrt(47))/2 ,x=-1/2-i(sqrt(47))/2
Find the zeros of (x+3)(x+4)=6x
The zeros of (x+3)(x+4)=6x are x=-1/2+i(sqrt(47))/2 ,x=-1/2-i(sqrt(47))/2