What are the solutions to the equation 5x^2+x-8=0 ?

The solutions to the equation 5x^2+x-8=0 are x=(-1+sqrt(161))/(10),x=(-1-sqrt(161))/(10)

Find the zeros of 5x^2+x-8=0

The zeros of 5x^2+x-8=0 are x=(-1+sqrt(161))/(10),x=(-1-sqrt(161))/(10)

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5 x^{2} + x - 8 = 0

x = \frac{- 1 + 161}{10}, x = \frac{- 1 - 161}{10}

Decimal Notation

x = 1.16885 \dots, x = - 1.36885 \dots

Solution steps

5 x^{2} + x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 ( - 8 )}{2 \cdot 5}

1^{2} - 4 \cdot 5 (- 8) = 161

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

Separate the solutions

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

x = \frac{- 1 + 161}{2 \cdot 5} : \frac{- 1 + 161}{10}

x = \frac{- 1 - 161}{2 \cdot 5} : \frac{- 1 - 161}{10}

The solutions to the quadratic equation are:

x = \frac{- 1 + 161}{10}, x = \frac{- 1 - 161}{10}

5 y^{2} - 25 y = 0

0.1 x^{2} - 30 x + 1760 = 0

1 8^{2} = 9^{2} + x^{2}

\frac{1}{2} x^{2} = 50

(2 x - 3)^{2} = 8 x

What are the solutions to the equation 5x^2+x-8=0 ?
The solutions to the equation 5x^2+x-8=0 are x=(-1+sqrt(161))/(10),x=(-1-sqrt(161))/(10)
Find the zeros of 5x^2+x-8=0
The zeros of 5x^2+x-8=0 are x=(-1+sqrt(161))/(10),x=(-1-sqrt(161))/(10)