What are the solutions to the equation -1/2 x^2=-6x+20 ?

The solutions to the equation -1/2 x^2=-6x+20 are x=6-2i,x=6+2i

Find the zeros of -1/2 x^2=-6x+20

The zeros of -1/2 x^2=-6x+20 are x=6-2i,x=6+2i

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- \frac{1}{2} x^{2} = - 6 x + 20

x = 6 - 2 i, x = 6 + 2 i

Solution steps

- \frac{1}{2} x^{2} = - 6 x + 20

Multiply both sides by

2

- x^{2} = - 12 x + 40

Move

40

to the left side

- x^{2} - 40 = - 12 x

Move

12 x

to the left side

- x^{2} - 40 + 12 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} + 12 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 1 ) ( - 40 )}{2 ( - 1 )}

Simplify

1 2^{2} - 4 (- 1) (- 40) : 4 i

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

Separate the solutions

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

x = \frac{- 12 + 4 i}{2 ( - 1 )} : 6 - 2 i

x = \frac{- 12 - 4 i}{2 ( - 1 )} : 6 + 2 i

The solutions to the quadratic equation are:

x = 6 - 2 i, x = 6 + 2 i

(3 x^{\frac{1}{2}}) (4 x^{\frac{2}{3}})

- 4 x + 3 = - 5

\frac{10}{12} \cdot \frac{6}{10}

- 2 y + 3 (4 - y) = 12 - 5 y

y (ab + 4) - 5 (4 + ab)

What are the solutions to the equation -1/2 x^2=-6x+20 ?
The solutions to the equation -1/2 x^2=-6x+20 are x=6-2i,x=6+2i
Find the zeros of -1/2 x^2=-6x+20
The zeros of -1/2 x^2=-6x+20 are x=6-2i,x=6+2i