What are the solutions to the equation 2x^2+5x=9 ?

The solutions to the equation 2x^2+5x=9 are x=(-5+sqrt(97))/4 ,x=(-5-sqrt(97))/4

Find the zeros of 2x^2+5x=9

The zeros of 2x^2+5x=9 are x=(-5+sqrt(97))/4 ,x=(-5-sqrt(97))/4

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2 x^{2} + 5 x = 9

x = \frac{- 5 + 97}{4}, x = \frac{- 5 - 97}{4}

Solution steps

2 x^{2} + 5 x = 9

Move

9

to the left side

2 x^{2} + 5 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 9) = 97

x_{1, 2} = \frac{- 5 \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 97}{2 \cdot 2}, x_{2} = \frac{- 5 - 97}{2 \cdot 2}

x = \frac{- 5 + 97}{2 \cdot 2} : \frac{- 5 + 97}{4}

x = \frac{- 5 - 97}{2 \cdot 2} : \frac{- 5 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 97}{4}, x = \frac{- 5 - 97}{4}

2 x = x + 3

24 + 54

(x^{\frac{1}{5}})^{2}

lo g_{25} (\frac{1}{25})

x + 96 = x + 96

What are the solutions to the equation 2x^2+5x=9 ?
The solutions to the equation 2x^2+5x=9 are x=(-5+sqrt(97))/4 ,x=(-5-sqrt(97))/4
Find the zeros of 2x^2+5x=9
The zeros of 2x^2+5x=9 are x=(-5+sqrt(97))/4 ,x=(-5-sqrt(97))/4