What are the solutions to the equation -2x^2+3x-5=0 ?

The solutions to the equation -2x^2+3x-5=0 are x= 3/4-i(sqrt(31))/4 ,x= 3/4+i(sqrt(31))/4

Find the zeros of -2x^2+3x-5=0

The zeros of -2x^2+3x-5=0 are x= 3/4-i(sqrt(31))/4 ,x= 3/4+i(sqrt(31))/4

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- 2 x^{2} + 3 x - 5 = 0

x = \frac{3}{4} - i \frac{31}{4}, x = \frac{3}{4} + i \frac{31}{4}

Solution steps

- 2 x^{2} + 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

Simplify

3^{2} - 4 (- 2) (- 5) : 31 i

x_{1, 2} = \frac{- 3 \pm 31 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 3 + 31 i}{2 ( - 2 )}, x_{2} = \frac{- 3 - 31 i}{2 ( - 2 )}

x = \frac{- 3 + 31 i}{2 ( - 2 )} : \frac{3}{4} - i \frac{31}{4}

x = \frac{- 3 - 31 i}{2 ( - 2 )} : \frac{3}{4} + i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} - i \frac{31}{4}, x = \frac{3}{4} + i \frac{31}{4}

r, V = \frac{1}{3} π r^{2} h

x^{2} - 12 x - 40 = 0

r, x^{2} + y^{2} = r^{2}

x^{2} - 4 x

r, v = π r^{2} h

What are the solutions to the equation -2x^2+3x-5=0 ?
The solutions to the equation -2x^2+3x-5=0 are x= 3/4-i(sqrt(31))/4 ,x= 3/4+i(sqrt(31))/4
Find the zeros of -2x^2+3x-5=0
The zeros of -2x^2+3x-5=0 are x= 3/4-i(sqrt(31))/4 ,x= 3/4+i(sqrt(31))/4