What are the solutions to the equation 3000=-35x^2+1250x-6500 ?

The solutions to the equation 3000=-35x^2+1250x-6500 are x=(5(25-sqrt(93)))/7 ,x=(5(25+sqrt(93)))/7

Find the zeros of 3000=-35x^2+1250x-6500

The zeros of 3000=-35x^2+1250x-6500 are x=(5(25-sqrt(93)))/7 ,x=(5(25+sqrt(93)))/7

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3000 = - 35 x^{2} + 1250 x - 6500

x = \frac{5 ( 25 - 93 )}{7}, x = \frac{5 ( 25 + 93 )}{7}

Decimal Notation

x = 10.96882 \dots, x = 24.74546 \dots

Solution steps

3000 = - 35 x^{2} + 1250 x - 6500

Switch sides

- 35 x^{2} + 1250 x - 6500 = 3000

Move

3000

to the left side

- 35 x^{2} + 1250 x - 9500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1250 \pm 125 0 ^{2} - 4 ( - 35 ) ( - 9500 )}{2 ( - 35 )}

125 0^{2} - 4 (- 35) (- 9500) = 5093

x_{1, 2} = \frac{- 1250 \pm 50 93}{2 ( - 35 )}

Separate the solutions

x_{1} = \frac{- 1250 + 50 93}{2 ( - 35 )}, x_{2} = \frac{- 1250 - 50 93}{2 ( - 35 )}

x = \frac{- 1250 + 50 93}{2 ( - 35 )} : \frac{5 ( 25 - 93 )}{7}

x = \frac{- 1250 - 50 93}{2 ( - 35 )} : \frac{5 ( 25 + 93 )}{7}

The solutions to the quadratic equation are:

x = \frac{5 ( 25 - 93 )}{7}, x = \frac{5 ( 25 + 93 )}{7}

p^{2} - 1 p = 0

24 x^{2} + 28 x + 4 = 0

20 x^{2} - 5 = 0

x^{2} = 6 x - 66

x = x^{2} - 2 x - 3

What are the solutions to the equation 3000=-35x^2+1250x-6500 ?
The solutions to the equation 3000=-35x^2+1250x-6500 are x=(5(25-sqrt(93)))/7 ,x=(5(25+sqrt(93)))/7
Find the zeros of 3000=-35x^2+1250x-6500
The zeros of 3000=-35x^2+1250x-6500 are x=(5(25-sqrt(93)))/7 ,x=(5(25+sqrt(93)))/7