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 \cdot 2 \cdot 4}{2 \cdot 2}

Simplify

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

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

Separate the solutions

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

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

x = \frac{- ( - 5 ) - 7 i}{2 \cdot 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}

8 - 11

\frac{\frac{5}{x} + \frac{1}{2 y}}{\frac{1}{2}}

x^{2} - 5 x - 24

(\frac{3 x}{2})^{2}

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

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