What are the solutions to the equation 2x^2+11x-5=0 ?

The solutions to the equation 2x^2+11x-5=0 are x=(-11+sqrt(161))/4 ,x=(-11-sqrt(161))/4

Find the zeros of 2x^2+11x-5=0

The zeros of 2x^2+11x-5=0 are x=(-11+sqrt(161))/4 ,x=(-11-sqrt(161))/4

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2 x^{2} + 11 x - 5 = 0

x = \frac{- 11 + 161}{4}, x = \frac{- 11 - 161}{4}

Decimal Notation

x = 0.42214 \dots, x = - 5.92214 \dots

Solution steps

2 x^{2} + 11 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

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

x_{1, 2} = \frac{- 11 \pm 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 161}{2 \cdot 2}, x_{2} = \frac{- 11 - 161}{2 \cdot 2}

x = \frac{- 11 + 161}{2 \cdot 2} : \frac{- 11 + 161}{4}

x = \frac{- 11 - 161}{2 \cdot 2} : \frac{- 11 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{- 11 + 161}{4}, x = \frac{- 11 - 161}{4}

∣ 5 x - 2 ∣ < 8

x^{2} + 20 x + 200 = 0

2 a^{2} - a - 15

5 x^{2} - 3 x - 14

\frac{1}{3} + \frac{1}{15}

What are the solutions to the equation 2x^2+11x-5=0 ?
The solutions to the equation 2x^2+11x-5=0 are x=(-11+sqrt(161))/4 ,x=(-11-sqrt(161))/4
Find the zeros of 2x^2+11x-5=0
The zeros of 2x^2+11x-5=0 are x=(-11+sqrt(161))/4 ,x=(-11-sqrt(161))/4