What are the solutions to the equation 3x^2-4=27-x ?

The solutions to the equation 3x^2-4=27-x are x=(-1+sqrt(373))/6 ,x=(-1-sqrt(373))/6

Find the zeros of 3x^2-4=27-x

The zeros of 3x^2-4=27-x are x=(-1+sqrt(373))/6 ,x=(-1-sqrt(373))/6

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3 x^{2} - 4 = 27 - x

x = \frac{- 1 + 373}{6}, x = \frac{- 1 - 373}{6}

Decimal Notation

x = 3.05220 \dots, x = - 3.38553 \dots

Solution steps

3 x^{2} - 4 = 27 - x

Move

x

to the left side

3 x^{2} - 4 + x = 27

Move

27

to the left side

3 x^{2} + x - 31 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 31 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 31) = 373

x_{1, 2} = \frac{- 1 \pm 373}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 373}{2 \cdot 3}, x_{2} = \frac{- 1 - 373}{2 \cdot 3}

x = \frac{- 1 + 373}{2 \cdot 3} : \frac{- 1 + 373}{6}

x = \frac{- 1 - 373}{2 \cdot 3} : \frac{- 1 - 373}{6}

The solutions to the quadratic equation are:

x = \frac{- 1 + 373}{6}, x = \frac{- 1 - 373}{6}

x, 2 x^{2} - x y + y^{2} = 8

X, I = X^{2} + 2 X + 4

x^{2} + 2 x + 2 = 0

2 (x - 3) (x + 1) = x (x - 1) + 2

6 x^{2} - 26 x + 24 = 0

What are the solutions to the equation 3x^2-4=27-x ?
The solutions to the equation 3x^2-4=27-x are x=(-1+sqrt(373))/6 ,x=(-1-sqrt(373))/6
Find the zeros of 3x^2-4=27-x
The zeros of 3x^2-4=27-x are x=(-1+sqrt(373))/6 ,x=(-1-sqrt(373))/6