What are the solutions to the equation x^2-94=-x-4x ?

The solutions to the equation x^2-94=-x-4x are x=(-5+sqrt(401))/2 ,x=(-5-sqrt(401))/2

Find the zeros of x^2-94=-x-4x

The zeros of x^2-94=-x-4x are x=(-5+sqrt(401))/2 ,x=(-5-sqrt(401))/2

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x^{2} - 94 = - x - 4 x

x = \frac{- 5 + 401}{2}, x = \frac{- 5 - 401}{2}

Decimal Notation

x = 7.51249 \dots, x = - 12.51249 \dots

Solution steps

x^{2} - 94 = - x - 4 x

Refine

x^{2} - 94 = - 5 x

Move

5 x

to the left side

x^{2} - 94 + 5 x = 0

Write in the standard form

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

x^{2} + 5 x - 94 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 94 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 94) = 401

x_{1, 2} = \frac{- 5 \pm 401}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 401}{2 \cdot 1}, x_{2} = \frac{- 5 - 401}{2 \cdot 1}

x = \frac{- 5 + 401}{2 \cdot 1} : \frac{- 5 + 401}{2}

x = \frac{- 5 - 401}{2 \cdot 1} : \frac{- 5 - 401}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 401}{2}, x = \frac{- 5 - 401}{2}

x^{2} + 4 x + 20

x^{2} = 4

5 x^{2} = 12 x + 4

h^{2} = 4^{2} + 3^{2}

- 2 x^{2} + 6 x - 4 + 6 x + 7 x^{2} - 9 - 6 x^{2} = 2 + 4 x

What are the solutions to the equation x^2-94=-x-4x ?
The solutions to the equation x^2-94=-x-4x are x=(-5+sqrt(401))/2 ,x=(-5-sqrt(401))/2
Find the zeros of x^2-94=-x-4x
The zeros of x^2-94=-x-4x are x=(-5+sqrt(401))/2 ,x=(-5-sqrt(401))/2