What are the solutions to the equation 2x^2-12x+20=0 ?

The solutions to the equation 2x^2-12x+20=0 are x=3+i,x=3-i

Find the zeros of 2x^2-12x+20=0

The zeros of 2x^2-12x+20=0 are x=3+i,x=3-i

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2 x^{2} - 12 x + 20 = 0

x = 3 + i, x = 3 - i

Solution steps

2 x^{2} - 12 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

Simplify

(- 12)^{2} - 4 \cdot 2 \cdot 20 : 4 i

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

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 i}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 4 i}{2 \cdot 2}

x = \frac{- ( - 12 ) + 4 i}{2 \cdot 2} : 3 + i

x = \frac{- ( - 12 ) - 4 i}{2 \cdot 2} : 3 - i

The solutions to the quadratic equation are:

x = 3 + i, x = 3 - i

\frac{2}{3} (p + 2) > \frac{5}{6} (p - 3)

5 \cdot 16

\frac{8}{2 7}

\frac{x}{2} = \frac{9}{14}

\frac{x}{9} - 1 = - 10

What are the solutions to the equation 2x^2-12x+20=0 ?
The solutions to the equation 2x^2-12x+20=0 are x=3+i,x=3-i
Find the zeros of 2x^2-12x+20=0
The zeros of 2x^2-12x+20=0 are x=3+i,x=3-i