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=(sqrt(10)-2)/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=(sqrt(10)-2)/2

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

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

Decimal Notation

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- ( - 4 ) - 2 10}{2 ( - 2 )} : \frac{10 - 2}{2}

The solutions to the quadratic equation are:

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

8^{2} + 6^{2} = c^{2}

7^{\frac{1}{5}} \cdot 4 9^{\frac{7}{5}}

216 x^{3} y^{2} z^{7}

\frac{10 !}{( 10 - 3 ) ! \cdot 3 !}

188 - 818

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=(sqrt(10)-2)/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=(sqrt(10)-2)/2