What are the solutions to the equation 2x^2+x-20=0 ?

The solutions to the equation 2x^2+x-20=0 are x=(-1+sqrt(161))/4 ,x=(-1-sqrt(161))/4

Find the zeros of 2x^2+x-20=0

The zeros of 2x^2+x-20=0 are x=(-1+sqrt(161))/4 ,x=(-1-sqrt(161))/4

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2 x^{2} + x - 20 = 0

x = \frac{- 1 + 161}{4}, x = \frac{- 1 - 161}{4}

Decimal Notation

x = 2.92214 \dots, x = - 3.42214 \dots

Solution steps

2 x^{2} + x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 20) = 161

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

Separate the solutions

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

x = \frac{- 1 + 161}{2 \cdot 2} : \frac{- 1 + 161}{4}

x = \frac{- 1 - 161}{2 \cdot 2} : \frac{- 1 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 161}{4}, x = \frac{- 1 - 161}{4}

t, s = \frac{1}{2} g t^{2} + v t

X^{2} - X - 2 = 0

4 t^{2} - 9 = 0

x^{2} + \frac{3}{2} x - \frac{1}{2} = 0

2 y^{2} - 7 y - 4 = 0

What are the solutions to the equation 2x^2+x-20=0 ?
The solutions to the equation 2x^2+x-20=0 are x=(-1+sqrt(161))/4 ,x=(-1-sqrt(161))/4
Find the zeros of 2x^2+x-20=0
The zeros of 2x^2+x-20=0 are x=(-1+sqrt(161))/4 ,x=(-1-sqrt(161))/4