What are the solutions to the equation 2x^2-2x-30=-2x^2-19 ?

The solutions to the equation 2x^2-2x-30=-2x^2-19 are x=(1+3sqrt(5))/4 ,x=(1-3sqrt(5))/4

Find the zeros of 2x^2-2x-30=-2x^2-19

The zeros of 2x^2-2x-30=-2x^2-19 are x=(1+3sqrt(5))/4 ,x=(1-3sqrt(5))/4

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2 x^{2} - 2 x - 30 = - 2 x^{2} - 19

x = \frac{1 + 3 5}{4}, x = \frac{1 - 3 5}{4}

Decimal Notation

x = 1.92705 \dots, x = - 1.42705 \dots

Solution steps

2 x^{2} - 2 x - 30 = - 2 x^{2} - 19

Move

19

to the left side

2 x^{2} - 2 x - 11 = - 2 x^{2}

Move

2 x^{2}

to the left side

4 x^{2} - 2 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 11 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 11) = 65

x_{1, 2} = \frac{- ( - 2 ) \pm 6 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6 5}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 6 5}{2 \cdot 4}

x = \frac{- ( - 2 ) + 6 5}{2 \cdot 4} : \frac{1 + 3 5}{4}

x = \frac{- ( - 2 ) - 6 5}{2 \cdot 4} : \frac{1 - 3 5}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 3 5}{4}, x = \frac{1 - 3 5}{4}

v^{2} - 4 v - 12 = 0

- x^{2} - x + 6 = 0

- y^{2} + 8 y = 0

- x^{2} - 3 x - 1 = 0

x (x - 4) = - 8

What are the solutions to the equation 2x^2-2x-30=-2x^2-19 ?
The solutions to the equation 2x^2-2x-30=-2x^2-19 are x=(1+3sqrt(5))/4 ,x=(1-3sqrt(5))/4
Find the zeros of 2x^2-2x-30=-2x^2-19
The zeros of 2x^2-2x-30=-2x^2-19 are x=(1+3sqrt(5))/4 ,x=(1-3sqrt(5))/4