What are the solutions to the equation 11x^2+28x+12=0 ?

The solutions to the equation 11x^2+28x+12=0 are x=-6/11 ,x=-2

Find the zeros of 11x^2+28x+12=0

The zeros of 11x^2+28x+12=0 are x=-6/11 ,x=-2

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11 x^{2} + 28 x + 12 = 0

x = - \frac{6}{11}, x = - 2

Decimal Notation

x = - 0.54545 \dots, x = - 2

Solution steps

11 x^{2} + 28 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 11 \cdot 12}{2 \cdot 11}

2 8^{2} - 4 \cdot 11 \cdot 12 = 16

x_{1, 2} = \frac{- 28 \pm 16}{2 \cdot 11}

Separate the solutions

x_{1} = \frac{- 28 + 16}{2 \cdot 11}, x_{2} = \frac{- 28 - 16}{2 \cdot 11}

x = \frac{- 28 + 16}{2 \cdot 11} : - \frac{6}{11}

x = \frac{- 28 - 16}{2 \cdot 11} : - 2

The solutions to the quadratic equation are:

x = - \frac{6}{11}, x = - 2

- 16 t^{2} + 32 t + 128 = 0

\frac{2 + 5 i}{3 + 4 i}

4 (x + 8)^{2} = 36

- \frac{1}{2} < - 1

(- \frac{1}{2})^{1}

What are the solutions to the equation 11x^2+28x+12=0 ?
The solutions to the equation 11x^2+28x+12=0 are x=-6/11 ,x=-2
Find the zeros of 11x^2+28x+12=0
The zeros of 11x^2+28x+12=0 are x=-6/11 ,x=-2