What are the solutions to the equation 13(x^2-x)=-2-2x^2 ?

The solutions to the equation 13(x^2-x)=-2-2x^2 are x= 2/3 ,x= 1/5

Find the zeros of 13(x^2-x)=-2-2x^2

The zeros of 13(x^2-x)=-2-2x^2 are x= 2/3 ,x= 1/5

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13 (x^{2} - x) = - 2 - 2 x^{2}

x = \frac{2}{3}, x = \frac{1}{5}

Decimal Notation

x = 0.66666 \dots, x = 0.2

Solution steps

13 (x^{2} - x) = - 2 - 2 x^{2}

Expand

13 (x^{2} - x) : 13 x^{2} - 13 x

13 x^{2} - 13 x = - 2 - 2 x^{2}

Move

2 x^{2}

to the left side

15 x^{2} - 13 x = - 2

Move

2

to the left side

15 x^{2} - 13 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 15 \cdot 2}{2 \cdot 15}

(- 13)^{2} - 4 \cdot 15 \cdot 2 = 7

x_{1, 2} = \frac{- ( - 13 ) \pm 7}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 7}{2 \cdot 15}, x_{2} = \frac{- ( - 13 ) - 7}{2 \cdot 15}

x = \frac{- ( - 13 ) + 7}{2 \cdot 15} : \frac{2}{3}

x = \frac{- ( - 13 ) - 7}{2 \cdot 15} : \frac{1}{5}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = \frac{1}{5}

x (x - 2) = 143

- 2 x^{2} - 28 x = 80

x^{2} + 20 x + 28 = 9

m^{2} + m - 1 = 0

3 x^{2} + 22 x + 7 = 0

What are the solutions to the equation 13(x^2-x)=-2-2x^2 ?
The solutions to the equation 13(x^2-x)=-2-2x^2 are x= 2/3 ,x= 1/5
Find the zeros of 13(x^2-x)=-2-2x^2
The zeros of 13(x^2-x)=-2-2x^2 are x= 2/3 ,x= 1/5