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

The solutions to the equation 2x-1=-x^2+x+3 are x=-(1+sqrt(17))/2 ,x=(sqrt(17)-1)/2

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

The zeros of 2x-1=-x^2+x+3 are x=-(1+sqrt(17))/2 ,x=(sqrt(17)-1)/2

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

x = - \frac{1 + 17}{2}, x = \frac{17 - 1}{2}

Decimal Notation

x = - 2.56155 \dots, x = 1.56155 \dots

Solution steps

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

Switch sides

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

Move

1

to the left side

- x^{2} + x + 4 = 2 x

Move

2 x

to the left side

- x^{2} - x + 4 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 4 = 17

x_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 17}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 17}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 17}{2 ( - 1 )} : - \frac{1 + 17}{2}

x = \frac{- ( - 1 ) - 17}{2 ( - 1 )} : \frac{17 - 1}{2}

The solutions to the quadratic equation are:

x = - \frac{1 + 17}{2}, x = \frac{17 - 1}{2}

25 x^{2} + 10 x = 0

4 x^{2} + 7 x - 8 = - 8 x - 4

2 (3 x + 4) = \frac{10 x + 30}{2} 210 x + 30

3 x^{2} + 12 x - 4 = 0

x^{2} + 6 x = 25

What are the solutions to the equation 2x-1=-x^2+x+3 ?
The solutions to the equation 2x-1=-x^2+x+3 are x=-(1+sqrt(17))/2 ,x=(sqrt(17)-1)/2
Find the zeros of 2x-1=-x^2+x+3
The zeros of 2x-1=-x^2+x+3 are x=-(1+sqrt(17))/2 ,x=(sqrt(17)-1)/2