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

The solutions to the equation 2x^2+x-1=2x+1 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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

The zeros of 2x^2+x-1=2x+1 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

Decimal Notation

x = 1.28077 \dots, x = - 0.78077 \dots

Solution steps

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

Move

1

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

x = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

\frac{5 !}{1 ! ( 5 - 1 ) !}

4 p \geq - 28

x^{2} - 11 x + 32 = 0

\frac{2}{x + 6} > \frac{- 3}{x - 3}

(24 3^{2})^{\frac{1}{10}}

What are the solutions to the equation 2x^2+x-1=2x+1 ?
The solutions to the equation 2x^2+x-1=2x+1 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4
Find the zeros of 2x^2+x-1=2x+1
The zeros of 2x^2+x-1=2x+1 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4