What are the solutions to the equation x^2-2x-28=-6x+4-6x+4 ?

The solutions to the equation x^2-2x-28=-6x+4-6x+4 are x=-5+sqrt(61),x=-5-sqrt(61)

Find the zeros of x^2-2x-28=-6x+4-6x+4

The zeros of x^2-2x-28=-6x+4-6x+4 are x=-5+sqrt(61),x=-5-sqrt(61)

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x^{2} - 2 x - 28 = - 6 x + 4 - 6 x + 4

x = - 5 + 61, x = - 5 - 61

Decimal Notation

x = 2.81024 \dots, x = - 12.81024 \dots

Solution steps

x^{2} - 2 x - 28 = - 6 x + 4 - 6 x + 4

Refine

x^{2} - 2 x - 28 = - 12 x + 8

Move

8

to the left side

x^{2} - 2 x - 36 = - 12 x

Move

12 x

to the left side

x^{2} + 10 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 36 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 36) = 261

x_{1, 2} = \frac{- 10 \pm 2 61}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 2 61}{2 \cdot 1}, x_{2} = \frac{- 10 - 2 61}{2 \cdot 1}

x = \frac{- 10 + 2 61}{2 \cdot 1} : - 5 + 61

x = \frac{- 10 - 2 61}{2 \cdot 1} : - 5 - 61

The solutions to the quadratic equation are:

x = - 5 + 61, x = - 5 - 61

100 (25) = x^{2}

x, 4 x^{2} - 4 x y + y^{2} - 6 x + 3 y + 2 = 0

0 = \frac{1}{2} x^{2} + 2 x + 6

8 x^{2} = 15 - 3 x

\frac{x ^{2} - 36}{8} + \frac{x ^{2} - 64}{6} = 14

What are the solutions to the equation x^2-2x-28=-6x+4-6x+4 ?
The solutions to the equation x^2-2x-28=-6x+4-6x+4 are x=-5+sqrt(61),x=-5-sqrt(61)
Find the zeros of x^2-2x-28=-6x+4-6x+4
The zeros of x^2-2x-28=-6x+4-6x+4 are x=-5+sqrt(61),x=-5-sqrt(61)