What are the solutions to the equation 14x^2+x-26=2x^2-4 ?

The solutions to the equation 14x^2+x-26=2x^2-4 are x=(-1+sqrt(1057))/(24),x=(-1-sqrt(1057))/(24)

Find the zeros of 14x^2+x-26=2x^2-4

The zeros of 14x^2+x-26=2x^2-4 are x=(-1+sqrt(1057))/(24),x=(-1-sqrt(1057))/(24)

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14 x^{2} + x - 26 = 2 x^{2} - 4

x = \frac{- 1 + 1057}{24}, x = \frac{- 1 - 1057}{24}

Decimal Notation

x = 1.31298 \dots, x = - 1.39631 \dots

Solution steps

14 x^{2} + x - 26 = 2 x^{2} - 4

Move

4

to the left side

14 x^{2} + x - 22 = 2 x^{2}

Move

2 x^{2}

to the left side

12 x^{2} + x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 12 ( - 22 )}{2 \cdot 12}

1^{2} - 4 \cdot 12 (- 22) = 1057

x_{1, 2} = \frac{- 1 \pm 1057}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 1 + 1057}{2 \cdot 12}, x_{2} = \frac{- 1 - 1057}{2 \cdot 12}

x = \frac{- 1 + 1057}{2 \cdot 12} : \frac{- 1 + 1057}{24}

x = \frac{- 1 - 1057}{2 \cdot 12} : \frac{- 1 - 1057}{24}

The solutions to the quadratic equation are:

x = \frac{- 1 + 1057}{24}, x = \frac{- 1 - 1057}{24}

(x + 9) (x + 1) = 0

10 x^{2} - 20 x - 88 = - 8

n^{2} + 6 n + 5 = 0

5 x^{2} + 5 x = - 1

x^{2} + 11 x = 28

What are the solutions to the equation 14x^2+x-26=2x^2-4 ?
The solutions to the equation 14x^2+x-26=2x^2-4 are x=(-1+sqrt(1057))/(24),x=(-1-sqrt(1057))/(24)
Find the zeros of 14x^2+x-26=2x^2-4
The zeros of 14x^2+x-26=2x^2-4 are x=(-1+sqrt(1057))/(24),x=(-1-sqrt(1057))/(24)