What are the solutions to the equation 105=x-2x^2 ?

The solutions to the equation 105=x-2x^2 are x= 1/4-i(sqrt(839))/4 ,x= 1/4+i(sqrt(839))/4

Find the zeros of 105=x-2x^2

The zeros of 105=x-2x^2 are x= 1/4-i(sqrt(839))/4 ,x= 1/4+i(sqrt(839))/4

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105 = x - 2 x^{2}

x = \frac{1}{4} - i \frac{839}{4}, x = \frac{1}{4} + i \frac{839}{4}

Solution steps

105 = x - 2 x^{2}

Switch sides

x - 2 x^{2} = 105

Move

105

to the left side

x - 2 x^{2} - 105 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 x^{2} + x - 105 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) ( - 105 )}{2 ( - 2 )}

Simplify

1^{2} - 4 (- 2) (- 105) : 839 i

x_{1, 2} = \frac{- 1 \pm 839 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 1 + 839 i}{2 ( - 2 )}, x_{2} = \frac{- 1 - 839 i}{2 ( - 2 )}

x = \frac{- 1 + 839 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{839}{4}

x = \frac{- 1 - 839 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{839}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} - i \frac{839}{4}, x = \frac{1}{4} + i \frac{839}{4}

x, x^{2} - 2 x - 24 = 0

(1 - x)^{2} + 1 = 0

x^{2} + 2 x + 0.5 = 0

x^{2} - 21 x + 80 = 0

4 x^{2} - 7 = 9

What are the solutions to the equation 105=x-2x^2 ?
The solutions to the equation 105=x-2x^2 are x= 1/4-i(sqrt(839))/4 ,x= 1/4+i(sqrt(839))/4
Find the zeros of 105=x-2x^2
The zeros of 105=x-2x^2 are x= 1/4-i(sqrt(839))/4 ,x= 1/4+i(sqrt(839))/4