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

The solutions to the equation 2x^2-5/2 x+2=0 are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8

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

The zeros of 2x^2-5/2 x+2=0 are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8

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2 x^{2} - \frac{5}{2} x + 2 = 0

x = \frac{5}{8} + i \frac{39}{8}, x = \frac{5}{8} - i \frac{39}{8}

Solution steps

2 x^{2} - \frac{5}{2} x + 2 = 0

Multiply both sides by

2

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 4 : 39 i

x_{1, 2} = \frac{- ( - 5 ) \pm 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 39 i}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 39 i}{2 \cdot 4}

x = \frac{- ( - 5 ) + 39 i}{2 \cdot 4} : \frac{5}{8} + i \frac{39}{8}

x = \frac{- ( - 5 ) - 39 i}{2 \cdot 4} : \frac{5}{8} - i \frac{39}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} + i \frac{39}{8}, x = \frac{5}{8} - i \frac{39}{8}

∣ x ∣ > 3

5 ∣ 5 x - 3 ∣ - 3 = 2

2 (2)

\frac{x y}{6} \cdot \frac{4 x}{16 - 3 x ^{2}}

3 x^{2} = - 54

What are the solutions to the equation 2x^2-5/2 x+2=0 ?
The solutions to the equation 2x^2-5/2 x+2=0 are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8
Find the zeros of 2x^2-5/2 x+2=0
The zeros of 2x^2-5/2 x+2=0 are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8