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

The solutions to the equation 2=-2x^2+3x are x= 3/4-i(sqrt(7))/4 ,x= 3/4+i(sqrt(7))/4

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

The zeros of 2=-2x^2+3x are x= 3/4-i(sqrt(7))/4 ,x= 3/4+i(sqrt(7))/4

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

x = \frac{3}{4} - i \frac{7}{4}, x = \frac{3}{4} + i \frac{7}{4}

Solution steps

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

Switch sides

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

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

3^{2} - 4 (- 2) (- 2) : 7 i

x_{1, 2} = \frac{- 3 \pm 7 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 3 + 7 i}{2 ( - 2 )}, x_{2} = \frac{- 3 - 7 i}{2 ( - 2 )}

x = \frac{- 3 + 7 i}{2 ( - 2 )} : \frac{3}{4} - i \frac{7}{4}

x = \frac{- 3 - 7 i}{2 ( - 2 )} : \frac{3}{4} + i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} - i \frac{7}{4}, x = \frac{3}{4} + i \frac{7}{4}

\frac{7 p + 4}{8} = 3

x (x + 9) = 0

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

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

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

What are the solutions to the equation 2=-2x^2+3x ?
The solutions to the equation 2=-2x^2+3x are x= 3/4-i(sqrt(7))/4 ,x= 3/4+i(sqrt(7))/4
Find the zeros of 2=-2x^2+3x
The zeros of 2=-2x^2+3x are x= 3/4-i(sqrt(7))/4 ,x= 3/4+i(sqrt(7))/4