What are the solutions to the equation 2x^2+1=2x+9 ?

The solutions to the equation 2x^2+1=2x+9 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

Find the zeros of 2x^2+1=2x+9

The zeros of 2x^2+1=2x+9 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2

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2 x^{2} + 1 = 2 x + 9

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

Decimal Notation

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

Solution steps

2 x^{2} + 1 = 2 x + 9

Move

9

to the left side

2 x^{2} - 8 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 8) = 217

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x - 4 = 8

(2 x - 9)^{2} = 9

\frac{\frac{7}{w} + \frac{3}{x}}{\frac{3}{w} - \frac{7}{x}}

(\frac{1}{3}) - 1

\frac{x}{2} - \frac{19}{6} \leq \frac{x}{3} + 1

What are the solutions to the equation 2x^2+1=2x+9 ?
The solutions to the equation 2x^2+1=2x+9 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2
Find the zeros of 2x^2+1=2x+9
The zeros of 2x^2+1=2x+9 are x=(1+sqrt(17))/2 ,x=(1-sqrt(17))/2