What are the solutions to the equation 2x^2-16x+40=0 ?

The solutions to the equation 2x^2-16x+40=0 are x=4+2i,x=4-2i

Find the zeros of 2x^2-16x+40=0

The zeros of 2x^2-16x+40=0 are x=4+2i,x=4-2i

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2 x^{2} - 16 x + 40 = 0

x = 4 + 2 i, x = 4 - 2 i

Solution steps

2 x^{2} - 16 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 40}{2 \cdot 2}

Simplify

(- 16)^{2} - 4 \cdot 2 \cdot 40 : 8 i

x_{1, 2} = \frac{- ( - 16 ) \pm 8 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8 i}{2 \cdot 2}, x_{2} = \frac{- ( - 16 ) - 8 i}{2 \cdot 2}

x = \frac{- ( - 16 ) + 8 i}{2 \cdot 2} : 4 + 2 i

x = \frac{- ( - 16 ) - 8 i}{2 \cdot 2} : 4 - 2 i

The solutions to the quadratic equation are:

x = 4 + 2 i, x = 4 - 2 i

1 2^{2} = (10 + x) (x)

b, a^{3} - a b^{2} - a c^{2} - ab c + b^{2} c + b c^{2} = 0

x^{2} - 13 + 12 = 0

8 x^{2} - 10 x + 2 = 0

x, (x - 3)^{2} + (y + 5)^{2} + z^{2} = 81

What are the solutions to the equation 2x^2-16x+40=0 ?
The solutions to the equation 2x^2-16x+40=0 are x=4+2i,x=4-2i
Find the zeros of 2x^2-16x+40=0
The zeros of 2x^2-16x+40=0 are x=4+2i,x=4-2i