What are the solutions to the equation 2x^2-16x+82=0 ?

The solutions to the equation 2x^2-16x+82=0 are x=4+5i,x=4-5i

Find the zeros of 2x^2-16x+82=0

The zeros of 2x^2-16x+82=0 are x=4+5i,x=4-5i

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2 x^{2} - 16 x + 82 = 0

x = 4 + 5 i, x = 4 - 5 i

Solution steps

2 x^{2} - 16 x + 82 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 82}{2 \cdot 2}

Simplify

(- 16)^{2} - 4 \cdot 2 \cdot 82 : 20 i

x_{1, 2} = \frac{- ( - 16 ) \pm 20 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 20 i}{2 \cdot 2}, x_{2} = \frac{- ( - 16 ) - 20 i}{2 \cdot 2}

x = \frac{- ( - 16 ) + 20 i}{2 \cdot 2} : 4 + 5 i

x = \frac{- ( - 16 ) - 20 i}{2 \cdot 2} : 4 - 5 i

The solutions to the quadratic equation are:

x = 4 + 5 i, x = 4 - 5 i

r^{2} + 8 = 0

2 x^{2} - 9 x + 7 = 0

x, (x + y)^{2} = 16

x, 3 z = - y^{2} + x^{2}

- 3 x^{2} + 24 x + 144 = 0

What are the solutions to the equation 2x^2-16x+82=0 ?
The solutions to the equation 2x^2-16x+82=0 are x=4+5i,x=4-5i
Find the zeros of 2x^2-16x+82=0
The zeros of 2x^2-16x+82=0 are x=4+5i,x=4-5i