What are the solutions to the equation -160x^2+450x-300=0 ?

The solutions to the equation -160x^2+450x-300=0 are x=(45-sqrt(105))/(32),x=(45+sqrt(105))/(32)

Find the zeros of -160x^2+450x-300=0

The zeros of -160x^2+450x-300=0 are x=(45-sqrt(105))/(32),x=(45+sqrt(105))/(32)

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- 160 x^{2} + 450 x - 300 = 0

x = \frac{45 - 105}{32}, x = \frac{45 + 105}{32}

Decimal Notation

x = 1.08603 \dots, x = 1.72646 \dots

Solution steps

- 160 x^{2} + 450 x - 300 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 450 \pm 45 0 ^{2} - 4 ( - 160 ) ( - 300 )}{2 ( - 160 )}

45 0^{2} - 4 (- 160) (- 300) = 10105

x_{1, 2} = \frac{- 450 \pm 10 105}{2 ( - 160 )}

Separate the solutions

x_{1} = \frac{- 450 + 10 105}{2 ( - 160 )}, x_{2} = \frac{- 450 - 10 105}{2 ( - 160 )}

x = \frac{- 450 + 10 105}{2 ( - 160 )} : \frac{45 - 105}{32}

x = \frac{- 450 - 10 105}{2 ( - 160 )} : \frac{45 + 105}{32}

The solutions to the quadratic equation are:

x = \frac{45 - 105}{32}, x = \frac{45 + 105}{32}

g (g + 10) = - 6

1000 x \cdot \frac{x}{2} = 5643 (104 - x)

9^{2} + 1 5^{2} = c^{2}

1200 - 280 x + 12 x^{2} = 0

8 k^{2} = 544

What are the solutions to the equation -160x^2+450x-300=0 ?
The solutions to the equation -160x^2+450x-300=0 are x=(45-sqrt(105))/(32),x=(45+sqrt(105))/(32)
Find the zeros of -160x^2+450x-300=0
The zeros of -160x^2+450x-300=0 are x=(45-sqrt(105))/(32),x=(45+sqrt(105))/(32)