What are the solutions to the equation 2x^2-12x+19=0 ?

The solutions to the equation 2x^2-12x+19=0 are x=3+i(sqrt(2))/2 ,x=3-i(sqrt(2))/2

Find the zeros of 2x^2-12x+19=0

The zeros of 2x^2-12x+19=0 are x=3+i(sqrt(2))/2 ,x=3-i(sqrt(2))/2

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2 x^{2} - 12 x + 19 = 0

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

Solution steps

2 x^{2} - 12 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 \cdot 19}{2 \cdot 2}

Simplify

(- 12)^{2} - 4 \cdot 2 \cdot 19 : 22 i

x_{1, 2} = \frac{- ( - 12 ) \pm 2 2 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 2 i}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 2 2 i}{2 \cdot 2}

x = \frac{- ( - 12 ) + 2 2 i}{2 \cdot 2} : 3 + i \frac{2}{2}

x = \frac{- ( - 12 ) - 2 2 i}{2 \cdot 2} : 3 - i \frac{2}{2}

The solutions to the quadratic equation are:

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

- 6 x^{2} + 90 x - 300 = 0

3 x^{(2)} + 6 x - 45 = 0

a^{2} - 4 a + 3 = 0

z^{2} - 2 z + 3 = 0

- x^{2} - 6 = 0

What are the solutions to the equation 2x^2-12x+19=0 ?
The solutions to the equation 2x^2-12x+19=0 are x=3+i(sqrt(2))/2 ,x=3-i(sqrt(2))/2
Find the zeros of 2x^2-12x+19=0
The zeros of 2x^2-12x+19=0 are x=3+i(sqrt(2))/2 ,x=3-i(sqrt(2))/2