What are the solutions to the equation x^2-14x-284=-2x ?

The solutions to the equation x^2-14x-284=-2x are x=2(3+4sqrt(5)),x=2(3-4sqrt(5))

Find the zeros of x^2-14x-284=-2x

The zeros of x^2-14x-284=-2x are x=2(3+4sqrt(5)),x=2(3-4sqrt(5))

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x^{2} - 14 x - 284 = - 2 x

x = 2 (3 + 45), x = 2 (3 - 45)

Solution steps

x^{2} - 14 x - 284 = - 2 x

Move

2 x

to the left side

x^{2} - 12 x - 284 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 284 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 284) = 165

x_{1, 2} = \frac{- ( - 12 ) \pm 16 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 16 5}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 16 5}{2 \cdot 1}

x = \frac{- ( - 12 ) + 16 5}{2 \cdot 1} : 2 (3 + 45)

x = \frac{- ( - 12 ) - 16 5}{2 \cdot 1} : 2 (3 - 45)

The solutions to the quadratic equation are:

x = 2 (3 + 45), x = 2 (3 - 45)

4 x^{2} - 12 x + 9 = 0

5^{2} + b^{2} = 9^{2}

3 y^{2} - 5 y + 2 = 0

r^{2} - r - 1 = 0

x^{2} - x = 9

What are the solutions to the equation x^2-14x-284=-2x ?
The solutions to the equation x^2-14x-284=-2x are x=2(3+4sqrt(5)),x=2(3-4sqrt(5))
Find the zeros of x^2-14x-284=-2x
The zeros of x^2-14x-284=-2x are x=2(3+4sqrt(5)),x=2(3-4sqrt(5))