What are the solutions to the equation -2x^2-5x+9=0 ?

The solutions to the equation -2x^2-5x+9=0 are x=-(5+sqrt(97))/4 ,x=(sqrt(97)-5)/4

Find the zeros of -2x^2-5x+9=0

The zeros of -2x^2-5x+9=0 are x=-(5+sqrt(97))/4 ,x=(sqrt(97)-5)/4

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

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

Decimal Notation

x = - 3.71221 \dots, x = 1.21221 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 42 a + 9 a^{2} + 49

5.2 q - 4.6 - 7.1 q = - 0.9 q - 4.6

3 x + 15 = x + 21

\frac{49 + 12 i}{22 - 5 i}

4 t^{2} + 9 = 15 t

What are the solutions to the equation -2x^2-5x+9=0 ?
The solutions to the equation -2x^2-5x+9=0 are x=-(5+sqrt(97))/4 ,x=(sqrt(97)-5)/4
Find the zeros of -2x^2-5x+9=0
The zeros of -2x^2-5x+9=0 are x=-(5+sqrt(97))/4 ,x=(sqrt(97)-5)/4