What are the solutions to the equation 2x^2-2x+25=0 ?

The solutions to the equation 2x^2-2x+25=0 are x= 1/2+i 7/2 ,x= 1/2-i 7/2

Find the zeros of 2x^2-2x+25=0

The zeros of 2x^2-2x+25=0 are x= 1/2+i 7/2 ,x= 1/2-i 7/2

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2 x^{2} - 2 x + 25 = 0

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

Solution steps

2 x^{2} - 2 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 25}{2 \cdot 2}

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 25 : 14 i

x_{1, 2} = \frac{- ( - 2 ) \pm 14 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 14 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 14 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 14 i}{2 \cdot 2} : \frac{1}{2} + i \frac{7}{2}

x = \frac{- ( - 2 ) - 14 i}{2 \cdot 2} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

7 - (3 + 4 i) + 6 i

- 5 \cdot (- 9)

\frac{3 π}{4} + \frac{3 π}{4}

s^{2} - 8 s - 9

∣ x + 4 ∣ \leq 4

What are the solutions to the equation 2x^2-2x+25=0 ?
The solutions to the equation 2x^2-2x+25=0 are x= 1/2+i 7/2 ,x= 1/2-i 7/2
Find the zeros of 2x^2-2x+25=0
The zeros of 2x^2-2x+25=0 are x= 1/2+i 7/2 ,x= 1/2-i 7/2