What are the solutions to the equation 2x^2-20x+50=-128 ?

The solutions to the equation 2x^2-20x+50=-128 are x=5+8i,x=5-8i

Find the zeros of 2x^2-20x+50=-128

The zeros of 2x^2-20x+50=-128 are x=5+8i,x=5-8i

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2 x^{2} - 20 x + 50 = - 128

x = 5 + 8 i, x = 5 - 8 i

Solution steps

2 x^{2} - 20 x + 50 = - 128

Move

128

to the left side

2 x^{2} - 20 x + 178 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 178}{2 \cdot 2}

Simplify

(- 20)^{2} - 4 \cdot 2 \cdot 178 : 32 i

x_{1, 2} = \frac{- ( - 20 ) \pm 32 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 32 i}{2 \cdot 2}, x_{2} = \frac{- ( - 20 ) - 32 i}{2 \cdot 2}

x = \frac{- ( - 20 ) + 32 i}{2 \cdot 2} : 5 + 8 i

x = \frac{- ( - 20 ) - 32 i}{2 \cdot 2} : 5 - 8 i

The solutions to the quadratic equation are:

x = 5 + 8 i, x = 5 - 8 i

x, 150 = x^{2} + 2 x + 1

- (x - 2)^{2} + 4 = - 5

4 x^{2} - 12 x + 16 = 0

x^{2} = 6 x - 7

y, z = 1 - y^{2}

What are the solutions to the equation 2x^2-20x+50=-128 ?
The solutions to the equation 2x^2-20x+50=-128 are x=5+8i,x=5-8i
Find the zeros of 2x^2-20x+50=-128
The zeros of 2x^2-20x+50=-128 are x=5+8i,x=5-8i