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

The solutions to the equation 2x^2+5x-5=0 are x=(-5+sqrt(65))/4 ,x=(-5-sqrt(65))/4

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

The zeros of 2x^2+5x-5=0 are x=(-5+sqrt(65))/4 ,x=(-5-sqrt(65))/4

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

x = \frac{- 5 + 65}{4}, x = \frac{- 5 - 65}{4}

Decimal Notation

x = 0.76556 \dots, x = - 3.26556 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 5) = 65

x_{1, 2} = \frac{- 5 \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 65}{2 \cdot 2}, x_{2} = \frac{- 5 - 65}{2 \cdot 2}

x = \frac{- 5 + 65}{2 \cdot 2} : \frac{- 5 + 65}{4}

x = \frac{- 5 - 65}{2 \cdot 2} : \frac{- 5 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 65}{4}, x = \frac{- 5 - 65}{4}

\frac{2}{2 5}

\frac{- 3}{5} + 7 \frac{1}{2} \leq \frac{- 7 x - 11}{2} + 3 < 8

\frac{1}{2} + \frac{3}{5}

- 1.8 \leq x \leq - 0.03

What are the solutions to the equation 2x^2+5x-5=0 ?
The solutions to the equation 2x^2+5x-5=0 are x=(-5+sqrt(65))/4 ,x=(-5-sqrt(65))/4
Find the zeros of 2x^2+5x-5=0
The zeros of 2x^2+5x-5=0 are x=(-5+sqrt(65))/4 ,x=(-5-sqrt(65))/4