What are the solutions to the equation -2x^2-5x+27=0 ?

The solutions to the equation -2x^2-5x+27=0 are x=-(5+sqrt(241))/4 ,x=(sqrt(241)-5)/4

Find the zeros of -2x^2-5x+27=0

The zeros of -2x^2-5x+27=0 are x=-(5+sqrt(241))/4 ,x=(sqrt(241)-5)/4

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- 2 x^{2} - 5 x + 27 = 0

x = - \frac{5 + 241}{4}, x = \frac{241 - 5}{4}

Solution steps

- 2 x^{2} - 5 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 27}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 27 = 241

x_{1, 2} = \frac{- ( - 5 ) \pm 241}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 241}{2 ( - 2 )}, x_{2} = \frac{- ( - 5 ) - 241}{2 ( - 2 )}

x = \frac{- ( - 5 ) + 241}{2 ( - 2 )} : - \frac{5 + 241}{4}

x = \frac{- ( - 5 ) - 241}{2 ( - 2 )} : \frac{241 - 5}{4}

The solutions to the quadratic equation are:

x = - \frac{5 + 241}{4}, x = \frac{241 - 5}{4}

0.2 x^{2} - 0.4 x - 15 = 0

x = 2 x^{2} + 5

- 0.2 x^{2} - 0.3 x + 1 = 0

- x^{2} - x = - 4

x (x + 7) = 14

What are the solutions to the equation -2x^2-5x+27=0 ?
The solutions to the equation -2x^2-5x+27=0 are x=-(5+sqrt(241))/4 ,x=(sqrt(241)-5)/4
Find the zeros of -2x^2-5x+27=0
The zeros of -2x^2-5x+27=0 are x=-(5+sqrt(241))/4 ,x=(sqrt(241)-5)/4