What are the solutions to the equation 5x^2=41-5x ?

The solutions to the equation 5x^2=41-5x are x=(-sqrt(5)+13)/(2sqrt(5)),x=-(sqrt(5)+13)/(2sqrt(5))

Find the zeros of 5x^2=41-5x

The zeros of 5x^2=41-5x are x=(-sqrt(5)+13)/(2sqrt(5)),x=-(sqrt(5)+13)/(2sqrt(5))

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5 x^{2} = 41 - 5 x

x = \frac{- 5 + 13}{2 5}, x = - \frac{5 + 13}{2 5}

Decimal Notation

x = 2.40688 \dots, x = - 3.40688 \dots

Solution steps

5 x^{2} = 41 - 5 x

Move

5 x

to the left side

5 x^{2} + 5 x = 41

Move

41

to the left side

5 x^{2} + 5 x - 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 5 ( - 41 )}{2 \cdot 5}

5^{2} - 4 \cdot 5 (- 41) = 135

x_{1, 2} = \frac{- 5 \pm 13 5}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 5 + 13 5}{2 \cdot 5}, x_{2} = \frac{- 5 - 13 5}{2 \cdot 5}

x = \frac{- 5 + 13 5}{2 \cdot 5} : \frac{- 5 + 13}{2 5}

x = \frac{- 5 - 13 5}{2 \cdot 5} : - \frac{5 + 13}{2 5}

The solutions to the quadratic equation are:

x = \frac{- 5 + 13}{2 5}, x = - \frac{5 + 13}{2 5}

\frac{1}{9} + \frac{2}{3}

8 r - 6 r + 2 = 5 r - r

x^{2} - 20 x

5 i - 10

- 1 + 5

What are the solutions to the equation 5x^2=41-5x ?
The solutions to the equation 5x^2=41-5x are x=(-sqrt(5)+13)/(2sqrt(5)),x=-(sqrt(5)+13)/(2sqrt(5))
Find the zeros of 5x^2=41-5x
The zeros of 5x^2=41-5x are x=(-sqrt(5)+13)/(2sqrt(5)),x=-(sqrt(5)+13)/(2sqrt(5))