What are the solutions to the equation 4x^2+12x+8=0 ?

The solutions to the equation 4x^2+12x+8=0 are x=-1,x=-2

Find the zeros of 4x^2+12x+8=0

The zeros of 4x^2+12x+8=0 are x=-1,x=-2

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4 x^{2} + 12 x + 8 = 0

x = - 1, x = - 2

Solution steps

4 x^{2} + 12 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 8}{2 \cdot 4}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = - 2

5 n^{2} - 7 = 488

2 (x - 3) = 2 x - 6

- 160 - 480 + 960 + \frac{12}{5} x = 0

∣ x - 4 ∣ > 7

n^{2} - 9 n + 18 = 0

What are the solutions to the equation 4x^2+12x+8=0 ?
The solutions to the equation 4x^2+12x+8=0 are x=-1,x=-2
Find the zeros of 4x^2+12x+8=0
The zeros of 4x^2+12x+8=0 are x=-1,x=-2