What are the solutions to the equation 6x^2+1=-8x ?

The solutions to the equation 6x^2+1=-8x are x=(-4+sqrt(10))/6 ,x=-(4+sqrt(10))/6

Find the zeros of 6x^2+1=-8x

The zeros of 6x^2+1=-8x are x=(-4+sqrt(10))/6 ,x=-(4+sqrt(10))/6

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6 x^{2} + 1 = - 8 x

x = \frac{- 4 + 10}{6}, x = - \frac{4 + 10}{6}

Decimal Notation

x = - 0.13962 \dots, x = - 1.19371 \dots

Solution steps

6 x^{2} + 1 = - 8 x

Move

8 x

to the left side

6 x^{2} + 1 + 8 x = 0

Write in the standard form

a x^{2} + b x + c = 0

6 x^{2} + 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

8^{2} - 4 \cdot 6 \cdot 1 = 210

x_{1, 2} = \frac{- 8 \pm 2 10}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 8 + 2 10}{2 \cdot 6}, x_{2} = \frac{- 8 - 2 10}{2 \cdot 6}

x = \frac{- 8 + 2 10}{2 \cdot 6} : \frac{- 4 + 10}{6}

x = \frac{- 8 - 2 10}{2 \cdot 6} : - \frac{4 + 10}{6}

The solutions to the quadratic equation are:

x = \frac{- 4 + 10}{6}, x = - \frac{4 + 10}{6}

2 X^{2} - 2 X + 1 = 0

P, P^{2} - P (\frac{1}{l} + 2 l) + 2 = 0

6 x^{2} + 480 = 0

x^{2} - 15 x + 12 = 0

x, x^{2} + 14 x + 49 = 0

What are the solutions to the equation 6x^2+1=-8x ?
The solutions to the equation 6x^2+1=-8x are x=(-4+sqrt(10))/6 ,x=-(4+sqrt(10))/6
Find the zeros of 6x^2+1=-8x
The zeros of 6x^2+1=-8x are x=(-4+sqrt(10))/6 ,x=-(4+sqrt(10))/6