What are the solutions to the equation 0.1x^2+36x+480=50x ?

The solutions to the equation 0.1x^2+36x+480=50x are x=80,x=60

Find the zeros of 0.1x^2+36x+480=50x

The zeros of 0.1x^2+36x+480=50x are x=80,x=60

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0.1 x^{2} + 36 x + 480 = 50 x

x = 80, x = 60

Solution steps

0.1 x^{2} + 36 x + 480 = 50 x

Multiply both sides by

10

x^{2} + 360 x + 4800 = 500 x

Move

500 x

to the left side

x^{2} - 140 x + 4800 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 1 \cdot 4800}{2 \cdot 1}

(- 140)^{2} - 4 \cdot 1 \cdot 4800 = 20

x_{1, 2} = \frac{- ( - 140 ) \pm 20}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 140 ) + 20}{2 \cdot 1}, x_{2} = \frac{- ( - 140 ) - 20}{2 \cdot 1}

x = \frac{- ( - 140 ) + 20}{2 \cdot 1} : 80

x = \frac{- ( - 140 ) - 20}{2 \cdot 1} : 60

The solutions to the quadratic equation are:

x = 80, x = 60

5 x^{2} - 25 x + 10 = 0

8 x^{3} - 1 = 8 x^{3} - 12 x^{2} + 6 x - 1

11 x^{2} - 33 x = 0

0 = x^{2} - 12

r^{2} + 4 r = - 1

What are the solutions to the equation 0.1x^2+36x+480=50x ?
The solutions to the equation 0.1x^2+36x+480=50x are x=80,x=60
Find the zeros of 0.1x^2+36x+480=50x
The zeros of 0.1x^2+36x+480=50x are x=80,x=60