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

The solutions to the equation 4(x^2+2x-4)=0 are x=-1+sqrt(5),x=-1-sqrt(5)

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

The zeros of 4(x^2+2x-4)=0 are x=-1+sqrt(5),x=-1-sqrt(5)

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

x = - 1 + 5, x = - 1 - 5

Solution steps

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

Expand

4 (x^{2} + 2 x - 4) : 4 x^{2} + 8 x - 16

4 x^{2} + 8 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 ( - 16 )}{2 \cdot 4}

8^{2} - 4 \cdot 4 (- 16) = 85

x_{1, 2} = \frac{- 8 \pm 8 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 8 + 8 5}{2 \cdot 4}, x_{2} = \frac{- 8 - 8 5}{2 \cdot 4}

x = \frac{- 8 + 8 5}{2 \cdot 4} : - 1 + 5

x = \frac{- 8 - 8 5}{2 \cdot 4} : - 1 - 5

The solutions to the quadratic equation are:

x = - 1 + 5, x = - 1 - 5

q (q + 7) = 5 q^{2} - 3

3 x^{2} - x^{2} + 7 x + 3 = 4

3 x^{2} - 6 x - 22 = 0

(x + 8) (x - 5) = 0

- 2 x^{2} + 4 x - 9 = 0

What are the solutions to the equation 4(x^2+2x-4)=0 ?
The solutions to the equation 4(x^2+2x-4)=0 are x=-1+sqrt(5),x=-1-sqrt(5)
Find the zeros of 4(x^2+2x-4)=0
The zeros of 4(x^2+2x-4)=0 are x=-1+sqrt(5),x=-1-sqrt(5)