What are the solutions to the equation -5x^2-9x+4=0 ?

The solutions to the equation -5x^2-9x+4=0 are x=-(9+sqrt(161))/(10),x=(sqrt(161)-9)/(10)

Find the zeros of -5x^2-9x+4=0

The zeros of -5x^2-9x+4=0 are x=-(9+sqrt(161))/(10),x=(sqrt(161)-9)/(10)

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- 5 x^{2} - 9 x + 4 = 0

x = - \frac{9 + 161}{10}, x = \frac{161 - 9}{10}

Decimal Notation

x = - 2.16885 \dots, x = 0.36885 \dots

Solution steps

- 5 x^{2} - 9 x + 4 = 0

Solve with the quadratic formula

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

(- 9)^{2} - 4 (- 5) \cdot 4 = 161

x_{1, 2} = \frac{- ( - 9 ) \pm 161}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 161}{2 ( - 5 )}, x_{2} = \frac{- ( - 9 ) - 161}{2 ( - 5 )}

x = \frac{- ( - 9 ) + 161}{2 ( - 5 )} : - \frac{9 + 161}{10}

x = \frac{- ( - 9 ) - 161}{2 ( - 5 )} : \frac{161 - 9}{10}

The solutions to the quadratic equation are:

x = - \frac{9 + 161}{10}, x = \frac{161 - 9}{10}

4 t^{2} + 12 t - 17 = 0

x^{2} - 9 = 0

2 s^{2} - 50 = 0

5 x^{2} - 6 x = 8

c = 1600 - 10 x + 0.25 x^{2}

What are the solutions to the equation -5x^2-9x+4=0 ?
The solutions to the equation -5x^2-9x+4=0 are x=-(9+sqrt(161))/(10),x=(sqrt(161)-9)/(10)
Find the zeros of -5x^2-9x+4=0
The zeros of -5x^2-9x+4=0 are x=-(9+sqrt(161))/(10),x=(sqrt(161)-9)/(10)