What are the solutions to the equation 0.2x^2-6x+80=0.6x^2-12x+80 ?

The solutions to the equation 0.2x^2-6x+80=0.6x^2-12x+80 are x=0,x=15

Find the zeros of 0.2x^2-6x+80=0.6x^2-12x+80

The zeros of 0.2x^2-6x+80=0.6x^2-12x+80 are x=0,x=15

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0.2 x^{2} - 6 x + 80 = 0.6 x^{2} - 12 x + 80

x = 0, x = 15

Solution steps

0.2 x^{2} - 6 x + 80 = 0.6 x^{2} - 12 x + 80

Multiply both sides by

10

2 x^{2} - 60 x + 800 = 6 x^{2} - 120 x + 800

Move

800

to the left side

2 x^{2} - 60 x = 6 x^{2} - 120 x

Move

120 x

to the left side

2 x^{2} + 60 x = 6 x^{2}

Move

6 x^{2}

to the left side

- 4 x^{2} + 60 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

6 0^{2} - 4 (- 4) \cdot 0 = 60

x_{1, 2} = \frac{- 60 \pm 60}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 60 + 60}{2 ( - 4 )}, x_{2} = \frac{- 60 - 60}{2 ( - 4 )}

x = \frac{- 60 + 60}{2 ( - 4 )} : 0

x = \frac{- 60 - 60}{2 ( - 4 )} : 15

The solutions to the quadratic equation are:

x = 0, x = 15

a, c = ba + a^{2}

- 2 x^{2} + 12 x - 8 = 0

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

(3 x + 2)^{2} - 5 (3 x + 2) = 0

b^{2} + 3 b - 10 = 0

What are the solutions to the equation 0.2x^2-6x+80=0.6x^2-12x+80 ?
The solutions to the equation 0.2x^2-6x+80=0.6x^2-12x+80 are x=0,x=15
Find the zeros of 0.2x^2-6x+80=0.6x^2-12x+80
The zeros of 0.2x^2-6x+80=0.6x^2-12x+80 are x=0,x=15