What are the solutions to the equation 2x^2-4x+12=0 ?

The solutions to the equation 2x^2-4x+12=0 are x=1+sqrt(5)i,x=1-sqrt(5)i

Find the zeros of 2x^2-4x+12=0

The zeros of 2x^2-4x+12=0 are x=1+sqrt(5)i,x=1-sqrt(5)i

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2 x^{2} - 4 x + 12 = 0

x = 1 + 5 i, x = 1 - 5 i

Solution steps

2 x^{2} - 4 x + 12 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 12 : 45 i

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 4 5 i}{2 \cdot 2} : 1 + 5 i

x = \frac{- ( - 4 ) - 4 5 i}{2 \cdot 2} : 1 - 5 i

The solutions to the quadratic equation are:

x = 1 + 5 i, x = 1 - 5 i

x^{3} - 4 x^{2} + 5 x - 2

3.4 a + 7.02 - 1.31 a = 9.2 a

3 \cdot 3

4 m - 9 = m + 7

8175

What are the solutions to the equation 2x^2-4x+12=0 ?
The solutions to the equation 2x^2-4x+12=0 are x=1+sqrt(5)i,x=1-sqrt(5)i
Find the zeros of 2x^2-4x+12=0
The zeros of 2x^2-4x+12=0 are x=1+sqrt(5)i,x=1-sqrt(5)i