What are the solutions to the equation 500x^2-2500x+35000=0 ?

The solutions to the equation 500x^2-2500x+35000=0 are x= 5/2+i(sqrt(255))/2 ,x= 5/2-i(sqrt(255))/2

Find the zeros of 500x^2-2500x+35000=0

The zeros of 500x^2-2500x+35000=0 are x= 5/2+i(sqrt(255))/2 ,x= 5/2-i(sqrt(255))/2

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500 x^{2} - 2500 x + 35000 = 0

x = \frac{5}{2} + i \frac{255}{2}, x = \frac{5}{2} - i \frac{255}{2}

Solution steps

500 x^{2} - 2500 x + 35000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2500 ) \pm ( - 2500 ) ^{2} - 4 \cdot 500 \cdot 35000}{2 \cdot 500}

Simplify

(- 2500)^{2} - 4 \cdot 500 \cdot 35000 : 500255 i

x_{1, 2} = \frac{- ( - 2500 ) \pm 500 255 i}{2 \cdot 500}

Separate the solutions

x_{1} = \frac{- ( - 2500 ) + 500 255 i}{2 \cdot 500}, x_{2} = \frac{- ( - 2500 ) - 500 255 i}{2 \cdot 500}

x = \frac{- ( - 2500 ) + 500 255 i}{2 \cdot 500} : \frac{5}{2} + i \frac{255}{2}

x = \frac{- ( - 2500 ) - 500 255 i}{2 \cdot 500} : \frac{5}{2} - i \frac{255}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{255}{2}, x = \frac{5}{2} - i \frac{255}{2}

0 = 6 x^{2} - 6

x^{2} = 9 x - 14

0 = - 4.9 t^{2} + 40 t

24 n^{2} + 6 n = 0

0 = - 16 t^{2} + 100 t + 23

What are the solutions to the equation 500x^2-2500x+35000=0 ?
The solutions to the equation 500x^2-2500x+35000=0 are x= 5/2+i(sqrt(255))/2 ,x= 5/2-i(sqrt(255))/2
Find the zeros of 500x^2-2500x+35000=0
The zeros of 500x^2-2500x+35000=0 are x= 5/2+i(sqrt(255))/2 ,x= 5/2-i(sqrt(255))/2