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

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

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

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

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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}

\frac{3}{5} + \frac{1}{4}

7 2^{2} + b^{2} = 7 5^{2}

y + 0.2 = 0.6 (y + 3)

\frac{\frac{2 x y ^{3} z}{x ^{2} - x z}}{\frac{4 y ^{2} z}{x - z}}

2 x^{2} + 12 x - 14 = 0

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