What are the solutions to the equation solvefor x,2x^2+4x-3=0 ?

The solutions to the equation solvefor x,2x^2+4x-3=0 are x=(-2+sqrt(10))/2 ,x=-(2+sqrt(10))/2

Find the zeros of solvefor x,2x^2+4x-3=0

The zeros of solvefor x,2x^2+4x-3=0 are x=(-2+sqrt(10))/2 ,x=-(2+sqrt(10))/2

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x, 2 x^{2} + 4 x - 3 = 0

x = \frac{- 2 + 10}{2}, x = - \frac{2 + 10}{2}

Decimal Notation

x = 0.58113 \dots, x = - 2.58113 \dots

Solution steps

2 x^{2} + 4 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 3) = 210

x_{1, 2} = \frac{- 4 \pm 2 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 10}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 10}{2 \cdot 2}

x = \frac{- 4 + 2 10}{2 \cdot 2} : \frac{- 2 + 10}{2}

x = \frac{- 4 - 2 10}{2 \cdot 2} : - \frac{2 + 10}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 10}{2}, x = - \frac{2 + 10}{2}

x, r an g o^{4} y^{2} = x^{2} - 4 x

- 2 q^{2} + 60 q - 120 = 0

5 x^{2} - 3 x = 7

3 y^{2} + 10 y - 8 = 0

(\frac{35}{6})^{2} + y^{2} = 1

What are the solutions to the equation solvefor x,2x^2+4x-3=0 ?
The solutions to the equation solvefor x,2x^2+4x-3=0 are x=(-2+sqrt(10))/2 ,x=-(2+sqrt(10))/2
Find the zeros of solvefor x,2x^2+4x-3=0
The zeros of solvefor x,2x^2+4x-3=0 are x=(-2+sqrt(10))/2 ,x=-(2+sqrt(10))/2