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

The solutions to the equation -2x^2+3x+4=0 are x=-(-3+sqrt(41))/4 ,x=(3+sqrt(41))/4

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

The zeros of -2x^2+3x+4=0 are x=-(-3+sqrt(41))/4 ,x=(3+sqrt(41))/4

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

x = - \frac{- 3 + 41}{4}, x = \frac{3 + 41}{4}

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 (- 2) \cdot 4 = 41

x_{1, 2} = \frac{- 3 \pm 41}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 3 + 41}{2 ( - 2 )}, x_{2} = \frac{- 3 - 41}{2 ( - 2 )}

x = \frac{- 3 + 41}{2 ( - 2 )} : - \frac{- 3 + 41}{4}

x = \frac{- 3 - 41}{2 ( - 2 )} : \frac{3 + 41}{4}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 41}{4}, x = \frac{3 + 41}{4}

48 = (2 x + 6) (2 x + 4)

x^{2} - 16 x = 21

x^{2} - 16 x = 20

x, x^{2} + 4 y^{2} - 4 x + 6 y + 20 = 0

X, Y = 2 X^{2}

What are the solutions to the equation -2x^2+3x+4=0 ?
The solutions to the equation -2x^2+3x+4=0 are x=-(-3+sqrt(41))/4 ,x=(3+sqrt(41))/4
Find the zeros of -2x^2+3x+4=0
The zeros of -2x^2+3x+4=0 are x=-(-3+sqrt(41))/4 ,x=(3+sqrt(41))/4