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

The solutions to the equation 0=2x^2-4x+5 are x=1+isqrt(3/2),x=1-isqrt(3/2)

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

The zeros of 0=2x^2-4x+5 are x=1+isqrt(3/2),x=1-isqrt(3/2)

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

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 5 : 26 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 6 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 6 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 6 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 6 i}{2 \cdot 2} : 1 + i \frac{3}{2}

x = \frac{- ( - 4 ) - 2 6 i}{2 \cdot 2} : 1 - i \frac{3}{2}

The solutions to the quadratic equation are:

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

5 (x - 2)^{2} - 500 = 0

x^{2} + 16 = - 28

8 = 4 + 17 t - 16 t^{2}

c^{2} - 5 c + 1 = 0

6 + 6 r^{2} = 15 r

What are the solutions to the equation 0=2x^2-4x+5 ?
The solutions to the equation 0=2x^2-4x+5 are x=1+isqrt(3/2),x=1-isqrt(3/2)
Find the zeros of 0=2x^2-4x+5
The zeros of 0=2x^2-4x+5 are x=1+isqrt(3/2),x=1-isqrt(3/2)