What are the solutions to the equation 4x^2+4(2x+1)=0 ?

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

Find the zeros of 4x^2+4(2x+1)=0

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

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4 x^{2} + 4 (2 x + 1) = 0

x = - 1

Solution steps

4 x^{2} + 4 (2 x + 1) = 0

Expand

4 x^{2} + 4 (2 x + 1) : 4 x^{2} + 8 x + 4

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 4 \cdot 4 = 0

x_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 4}

x = \frac{- 8}{2 \cdot 4}

\frac{- 8}{2 \cdot 4} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

- x^{2} - 56 = 0

0 = - 3 x^{2} + 2 x + 1

- 2.1 y^{2} + 8.04 y = 7.4

A, A^{2} + 2 A B + B^{2} = 100

4 r^{2} + 13 r + 3 = 0

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