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

The solutions to the equation 2x^2+2x-5=0 are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2

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

The zeros of 2x^2+2x-5=0 are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2

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

x = \frac{- 1 + 11}{2}, x = - \frac{1 + 11}{2}

Decimal Notation

x = 1.15831 \dots, x = - 2.15831 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 5) = 211

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

Separate the solutions

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

x = \frac{- 2 + 2 11}{2 \cdot 2} : \frac{- 1 + 11}{2}

x = \frac{- 2 - 2 11}{2 \cdot 2} : - \frac{1 + 11}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 11}{2}, x = - \frac{1 + 11}{2}

(8 a^{- 3} b^{9})^{\frac{2}{3}}

9 x^{2} = 36

\frac{2 ^{3}}{2 ^{- 4}}

∣ 2 x + 3 ∣ = 2 x + 3

35 - 25

What are the solutions to the equation 2x^2+2x-5=0 ?
The solutions to the equation 2x^2+2x-5=0 are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2
Find the zeros of 2x^2+2x-5=0
The zeros of 2x^2+2x-5=0 are x=(-1+sqrt(11))/2 ,x=-(1+sqrt(11))/2