What are the solutions to the equation -x^2+41x=395 ?

The solutions to the equation -x^2+41x=395 are x=(41-sqrt(101))/2 ,x=(41+sqrt(101))/2

Find the zeros of -x^2+41x=395

The zeros of -x^2+41x=395 are x=(41-sqrt(101))/2 ,x=(41+sqrt(101))/2

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- x^{2} + 41 x = 395

x = \frac{41 - 101}{2}, x = \frac{41 + 101}{2}

Decimal Notation

x = 15.47506 \dots, x = 25.52493 \dots

Solution steps

- x^{2} + 41 x = 395

Move

395

to the left side

- x^{2} + 41 x - 395 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 41 \pm 4 1 ^{2} - 4 ( - 1 ) ( - 395 )}{2 ( - 1 )}

4 1^{2} - 4 (- 1) (- 395) = 101

x_{1, 2} = \frac{- 41 \pm 101}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 41 + 101}{2 ( - 1 )}, x_{2} = \frac{- 41 - 101}{2 ( - 1 )}

x = \frac{- 41 + 101}{2 ( - 1 )} : \frac{41 - 101}{2}

x = \frac{- 41 - 101}{2 ( - 1 )} : \frac{41 + 101}{2}

The solutions to the quadratic equation are:

x = \frac{41 - 101}{2}, x = \frac{41 + 101}{2}

- x = 2 x

4 x > 24

36 v^{3} - 42 v^{2} + 42 v - 49

\frac{2}{3} (1)^{\frac{3}{2}}

44 v + 52 - v = - 77

What are the solutions to the equation -x^2+41x=395 ?
The solutions to the equation -x^2+41x=395 are x=(41-sqrt(101))/2 ,x=(41+sqrt(101))/2
Find the zeros of -x^2+41x=395
The zeros of -x^2+41x=395 are x=(41-sqrt(101))/2 ,x=(41+sqrt(101))/2