What are the solutions to the equation 3x(x+4)+(x+4)^2=0 ?

The solutions to the equation 3x(x+4)+(x+4)^2=0 are x=-1,x=-4

Find the zeros of 3x(x+4)+(x+4)^2=0

The zeros of 3x(x+4)+(x+4)^2=0 are x=-1,x=-4

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

x = - 1, x = - 4

Solution steps

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

Expand

3 x (x + 4) + (x + 4)^{2} : 4 x^{2} + 20 x + 16

4 x^{2} + 20 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 16 = 12

x_{1, 2} = \frac{- 20 \pm 12}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 12}{2 \cdot 4}, x_{2} = \frac{- 20 - 12}{2 \cdot 4}

x = \frac{- 20 + 12}{2 \cdot 4} : - 1

x = \frac{- 20 - 12}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = - 1, x = - 4

6 y^{2} + 4 y

4 x + x^{2} = 5

(x - 4) (x + 2) = 27

y^{2} - y - 20 = 0

- 3 x^{2} = 24 x + 48

What are the solutions to the equation 3x(x+4)+(x+4)^2=0 ?
The solutions to the equation 3x(x+4)+(x+4)^2=0 are x=-1,x=-4
Find the zeros of 3x(x+4)+(x+4)^2=0
The zeros of 3x(x+4)+(x+4)^2=0 are x=-1,x=-4