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

The solutions to the equation 2x^2+2x-7=0 are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2

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

The zeros of 2x^2+2x-7=0 are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2

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

x = \frac{- 1 + 15}{2}, x = - \frac{1 + 15}{2}

Decimal Notation

x = 1.43649 \dots, x = - 2.43649 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 7) = 215

x_{1, 2} = \frac{- 2 \pm 2 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 15}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 15}{2 \cdot 2}

x = \frac{- 2 + 2 15}{2 \cdot 2} : \frac{- 1 + 15}{2}

x = \frac{- 2 - 2 15}{2 \cdot 2} : - \frac{1 + 15}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 15}{2}, x = - \frac{1 + 15}{2}

- 47 \cdot 42

(16)^{\frac{5}{2}}

(\frac{3}{4})^{2} - \frac{1}{4} - (- \frac{3}{8})^{2} + (\frac{1}{4})^{3}

\frac{1}{t 1 + t} - \frac{1}{t}

6 v + 2 = 9 v + 14

What are the solutions to the equation 2x^2+2x-7=0 ?
The solutions to the equation 2x^2+2x-7=0 are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2
Find the zeros of 2x^2+2x-7=0
The zeros of 2x^2+2x-7=0 are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2