What are the solutions to the equation -2x^2+5x-4=0 ?

The solutions to the equation -2x^2+5x-4=0 are x= 5/4-i(sqrt(7))/4 ,x= 5/4+i(sqrt(7))/4

Find the zeros of -2x^2+5x-4=0

The zeros of -2x^2+5x-4=0 are x= 5/4-i(sqrt(7))/4 ,x= 5/4+i(sqrt(7))/4

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- 2 x^{2} + 5 x - 4 = 0

x = \frac{5}{4} - i \frac{7}{4}, x = \frac{5}{4} + i \frac{7}{4}

Solution steps

- 2 x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

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

Simplify

5^{2} - 4 (- 2) (- 4) : 7 i

x_{1, 2} = \frac{- 5 \pm 7 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 5 + 7 i}{2 ( - 2 )}, x_{2} = \frac{- 5 - 7 i}{2 ( - 2 )}

x = \frac{- 5 + 7 i}{2 ( - 2 )} : \frac{5}{4} - i \frac{7}{4}

x = \frac{- 5 - 7 i}{2 ( - 2 )} : \frac{5}{4} + i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} - i \frac{7}{4}, x = \frac{5}{4} + i \frac{7}{4}

1 - x - x^{2}

j^{2} + 8 j + 12

(7 + 2 i) (9 - 6 i)

\frac{2 a ^{3} b}{2 a b ^{3}}

5 x^{2} - 4 x - 1

What are the solutions to the equation -2x^2+5x-4=0 ?
The solutions to the equation -2x^2+5x-4=0 are x= 5/4-i(sqrt(7))/4 ,x= 5/4+i(sqrt(7))/4
Find the zeros of -2x^2+5x-4=0
The zeros of -2x^2+5x-4=0 are x= 5/4-i(sqrt(7))/4 ,x= 5/4+i(sqrt(7))/4