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

The solutions to the equation 2x^2+x-4=0,x^2+x=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4

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

The zeros of 2x^2+x-4=0,x^2+x=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4

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2 x^{2} + x - 4 = 0, x^{2} + x = 0

x = \frac{- 1 + 33}{4}, x = \frac{- 1 - 33}{4}

Decimal Notation

x = 1.18614 \dots, x = - 1.68614 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 4) = 33

x_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 33}{2 \cdot 2}, x_{2} = \frac{- 1 - 33}{2 \cdot 2}

x = \frac{- 1 + 33}{2 \cdot 2} : \frac{- 1 + 33}{4}

x = \frac{- 1 - 33}{2 \cdot 2} : \frac{- 1 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 33}{4}, x = \frac{- 1 - 33}{4}

3 x^{2} - 72 x + 324 = 0

q, (\frac{1}{2} q^{2} + 20) x = (86 - 6 q - 3 q^{2}) x

x, (z - 3)^{2} + (x - 2)^{2} = 25

2 1^{2} + 2 9^{2} = c^{2}

3 x^{2} + 8 x + 40 = 0

What are the solutions to the equation 2x^2+x-4=0,x^2+x=0 ?
The solutions to the equation 2x^2+x-4=0,x^2+x=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4
Find the zeros of 2x^2+x-4=0,x^2+x=0
The zeros of 2x^2+x-4=0,x^2+x=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4