What are the solutions to the equation 4x^2+12x=-1 ?

The solutions to the equation 4x^2+12x=-1 are x=(-3+2sqrt(2))/2 ,x=-(3+2sqrt(2))/2

Find the zeros of 4x^2+12x=-1

The zeros of 4x^2+12x=-1 are x=(-3+2sqrt(2))/2 ,x=-(3+2sqrt(2))/2

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4 x^{2} + 12 x = - 1

x = \frac{- 3 + 2 2}{2}, x = - \frac{3 + 2 2}{2}

Decimal Notation

x = - 0.08578 \dots, x = - 2.91421 \dots

Solution steps

4 x^{2} + 12 x = - 1

Move

1

to the left side

4 x^{2} + 12 x + 1 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 4 \cdot 1 = 82

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

Separate the solutions

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

x = \frac{- 12 + 8 2}{2 \cdot 4} : \frac{- 3 + 2 2}{2}

x = \frac{- 12 - 8 2}{2 \cdot 4} : - \frac{3 + 2 2}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 2 2}{2}, x = - \frac{3 + 2 2}{2}

x^{2} - 22 x + 4 = 0

- 16 t^{2} - 32 t + 1395 = 0

k^{2} - 6 = 0

2 r^{2} + 13 r + 17 = 0

180 = (x + 20) x 128

What are the solutions to the equation 4x^2+12x=-1 ?
The solutions to the equation 4x^2+12x=-1 are x=(-3+2sqrt(2))/2 ,x=-(3+2sqrt(2))/2
Find the zeros of 4x^2+12x=-1
The zeros of 4x^2+12x=-1 are x=(-3+2sqrt(2))/2 ,x=-(3+2sqrt(2))/2