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

The solutions to the equation 2x^2-1=9x are x=(9+sqrt(89))/4 ,x=(9-sqrt(89))/4

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

The zeros of 2x^2-1=9x are x=(9+sqrt(89))/4 ,x=(9-sqrt(89))/4

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2 x^{2} - 1 = 9 x

x = \frac{9 + 89}{4}, x = \frac{9 - 89}{4}

Solution steps

2 x^{2} - 1 = 9 x

Move

9 x

to the left side

2 x^{2} - 1 - 9 x = 0

Write in the standard form

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

2 x^{2} - 9 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 (- 1) = 89

x_{1, 2} = \frac{- ( - 9 ) \pm 89}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 89}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 89}{2 \cdot 2}

x = \frac{- ( - 9 ) + 89}{2 \cdot 2} : \frac{9 + 89}{4}

x = \frac{- ( - 9 ) - 89}{2 \cdot 2} : \frac{9 - 89}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 89}{4}, x = \frac{9 - 89}{4}

x^{2} = - 6 x + 7

x^{2} - 8 x = - 16

(p - q) (p - q)

(\infty - 3) \infty < 4 (\infty + 2)

54 g + 71 \leq - 37

What are the solutions to the equation 2x^2-1=9x ?
The solutions to the equation 2x^2-1=9x are x=(9+sqrt(89))/4 ,x=(9-sqrt(89))/4
Find the zeros of 2x^2-1=9x
The zeros of 2x^2-1=9x are x=(9+sqrt(89))/4 ,x=(9-sqrt(89))/4