What are the solutions to the equation 2x^2+7x+4=0 ?

The solutions to the equation 2x^2+7x+4=0 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4

Find the zeros of 2x^2+7x+4=0

The zeros of 2x^2+7x+4=0 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4

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2 x^{2} + 7 x + 4 = 0

x = \frac{- 7 + 17}{4}, x = \frac{- 7 - 17}{4}

Decimal Notation

x = - 0.71922 \dots, x = - 2.78077 \dots

Solution steps

2 x^{2} + 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 4 = 17

x_{1, 2} = \frac{- 7 \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 17}{2 \cdot 2}, x_{2} = \frac{- 7 - 17}{2 \cdot 2}

x = \frac{- 7 + 17}{2 \cdot 2} : \frac{- 7 + 17}{4}

x = \frac{- 7 - 17}{2 \cdot 2} : \frac{- 7 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 17}{4}, x = \frac{- 7 - 17}{4}

x^{2} + 7 x - 17 \leq 1

2 \cdot (x - 5) = 12

x^{6} - 4 x^{4} - 8 x^{4} + 32 x^{2} + 16 x^{2} - 64

- 10 ∣ v + 2 ∣ = 70

What are the solutions to the equation 2x^2+7x+4=0 ?
The solutions to the equation 2x^2+7x+4=0 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4
Find the zeros of 2x^2+7x+4=0
The zeros of 2x^2+7x+4=0 are x=(-7+sqrt(17))/4 ,x=(-7-sqrt(17))/4