What are the solutions to the equation x^2+1.2x=0.6x^2+1 ?

The solutions to the equation x^2+1.2x=0.6x^2+1 are x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2

Find the zeros of x^2+1.2x=0.6x^2+1

The zeros of x^2+1.2x=0.6x^2+1 are x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2

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x^{2} + 1.2 x = 0.6 x^{2} + 1

x = \frac{- 3 + 19}{2}, x = - \frac{3 + 19}{2}

Solution steps

x^{2} + 1.2 x = 0.6 x^{2} + 1

Multiply both sides by

10

10 x^{2} + 12 x = 6 x^{2} + 10

Move

10

to the left side

10 x^{2} + 12 x - 10 = 6 x^{2}

Move

6 x^{2}

to the left side

4 x^{2} + 12 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 10 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 10) = 419

x_{1, 2} = \frac{- 12 \pm 4 19}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 4 19}{2 \cdot 4}, x_{2} = \frac{- 12 - 4 19}{2 \cdot 4}

x = \frac{- 12 + 4 19}{2 \cdot 4} : \frac{- 3 + 19}{2}

x = \frac{- 12 - 4 19}{2 \cdot 4} : - \frac{3 + 19}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 19}{2}, x = - \frac{3 + 19}{2}

x^{2} + 14 x y + 49 y^{2}

2 = \frac{3 x + 14}{4}

x^{\frac{3}{2}} \cdot (x - \frac{1}{x})

x^{2} + x y - 20 y^{2}

5 x^{2} + 9 x = - 4

What are the solutions to the equation x^2+1.2x=0.6x^2+1 ?
The solutions to the equation x^2+1.2x=0.6x^2+1 are x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2
Find the zeros of x^2+1.2x=0.6x^2+1
The zeros of x^2+1.2x=0.6x^2+1 are x=(-3+sqrt(19))/2 ,x=-(3+sqrt(19))/2