What are the solutions to the equation -2x^2+20x-68=0 ?

The solutions to the equation -2x^2+20x-68=0 are x=5-3i,x=5+3i

Find the zeros of -2x^2+20x-68=0

The zeros of -2x^2+20x-68=0 are x=5-3i,x=5+3i

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- 2 x^{2} + 20 x - 68 = 0

x = 5 - 3 i, x = 5 + 3 i

Solution steps

- 2 x^{2} + 20 x - 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 2 ) ( - 68 )}{2 ( - 2 )}

Simplify

2 0^{2} - 4 (- 2) (- 68) : 12 i

x_{1, 2} = \frac{- 20 \pm 12 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 20 + 12 i}{2 ( - 2 )}, x_{2} = \frac{- 20 - 12 i}{2 ( - 2 )}

x = \frac{- 20 + 12 i}{2 ( - 2 )} : 5 - 3 i

x = \frac{- 20 - 12 i}{2 ( - 2 )} : 5 + 3 i

The solutions to the quadratic equation are:

x = 5 - 3 i, x = 5 + 3 i

x, x^{2} - 6 x y - 7 y^{2} + 10 x + 2 y + 17 = 0

(x + 1) (x - 1) = 2 (x^{2} - 13)

x^{2} + 2 x - 15 = 0

- x^{2} + 9 x - 8 = 0

2 (2 x^{2} - 5) = 3 (2 - 3 x^{2})

What are the solutions to the equation -2x^2+20x-68=0 ?
The solutions to the equation -2x^2+20x-68=0 are x=5-3i,x=5+3i
Find the zeros of -2x^2+20x-68=0
The zeros of -2x^2+20x-68=0 are x=5-3i,x=5+3i