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

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

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

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

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

x = - 1, x = - 2

Solution steps

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

Switch sides

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

18 x^{2} y + 60 x y^{2} + 50 y^{3}

∣ 9 ∣

- 2 - \frac{3}{2}

2^{l o g_{2} (n) + 1}

6 4^{\frac{5}{6}}

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