What are the solutions to the equation 3x^2+56=x^2-24x ?

The solutions to the equation 3x^2+56=x^2-24x are x=2(sqrt(2)-3),x=-2(3+sqrt(2))

Find the zeros of 3x^2+56=x^2-24x

The zeros of 3x^2+56=x^2-24x are x=2(sqrt(2)-3),x=-2(3+sqrt(2))

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3 x^{2} + 56 = x^{2} - 24 x

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

Decimal Notation

x = - 3.17157 \dots, x = - 8.82842 \dots

Solution steps

3 x^{2} + 56 = x^{2} - 24 x

Move

24 x

to the left side

3 x^{2} + 56 + 24 x = x^{2}

Move

x^{2}

to the left side

2 x^{2} + 24 x + 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 2 \cdot 56}{2 \cdot 2}

2 4^{2} - 4 \cdot 2 \cdot 56 = 82

x_{1, 2} = \frac{- 24 \pm 8 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 24 + 8 2}{2 \cdot 2}, x_{2} = \frac{- 24 - 8 2}{2 \cdot 2}

x = \frac{- 24 + 8 2}{2 \cdot 2} : 2 (2 - 3)

x = \frac{- 24 - 8 2}{2 \cdot 2} : - 2 (3 + 2)

The solutions to the quadratic equation are:

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

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

p^{2} - 100 = 0

- 28 q = - 96 q^{2} - 2

- 3 x^{2} - 30 = 0

x, e^{y} = x^{2} + y^{2}

What are the solutions to the equation 3x^2+56=x^2-24x ?
The solutions to the equation 3x^2+56=x^2-24x are x=2(sqrt(2)-3),x=-2(3+sqrt(2))
Find the zeros of 3x^2+56=x^2-24x
The zeros of 3x^2+56=x^2-24x are x=2(sqrt(2)-3),x=-2(3+sqrt(2))