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=(-9-sqrt(161))/(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=(-9-sqrt(161))/(10)

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

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

Decimal Notation

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 20 x^{2} + x + 12

\frac{3 x + 2}{2} = 5

lo g_{7} (9) = 2 x

\frac{ln ( 4 )}{ln ( 2 )}

11 \leq 2 x + 5 \leq 17

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=(-9-sqrt(161))/(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=(-9-sqrt(161))/(10)