What are the solutions to the equation 3x-5x^2+1=x-7 ?

The solutions to the equation 3x-5x^2+1=x-7 are x=-(-1+sqrt(41))/5 ,x=(1+sqrt(41))/5

Find the zeros of 3x-5x^2+1=x-7

The zeros of 3x-5x^2+1=x-7 are x=-(-1+sqrt(41))/5 ,x=(1+sqrt(41))/5

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3 x - 5 x^{2} + 1 = x - 7

x = - \frac{- 1 + 41}{5}, x = \frac{1 + 41}{5}

Decimal Notation

x = - 1.08062 \dots, x = 1.48062 \dots

Solution steps

3 x - 5 x^{2} + 1 = x - 7

Move

7

to the left side

- 5 x^{2} + 3 x + 8 = x

Move

x

to the left side

- 5 x^{2} + 2 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 8}{2 ( - 5 )}

2^{2} - 4 (- 5) \cdot 8 = 241

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

Separate the solutions

x_{1} = \frac{- 2 + 2 41}{2 ( - 5 )}, x_{2} = \frac{- 2 - 2 41}{2 ( - 5 )}

x = \frac{- 2 + 2 41}{2 ( - 5 )} : - \frac{- 1 + 41}{5}

x = \frac{- 2 - 2 41}{2 ( - 5 )} : \frac{1 + 41}{5}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 41}{5}, x = \frac{1 + 41}{5}

5 a^{2} - 3 a + 6 = 2

0 = 7 x^{2} - 2 x - 1

a, a^{2} + 2 ab - b^{2} = 0

t^{2} - 4.5 t + 3.5 = 0

6 x^{2} + 3 x = 0

What are the solutions to the equation 3x-5x^2+1=x-7 ?
The solutions to the equation 3x-5x^2+1=x-7 are x=-(-1+sqrt(41))/5 ,x=(1+sqrt(41))/5
Find the zeros of 3x-5x^2+1=x-7
The zeros of 3x-5x^2+1=x-7 are x=-(-1+sqrt(41))/5 ,x=(1+sqrt(41))/5