What are the solutions to the equation 2x^2-32x+178=0 ?

The solutions to the equation 2x^2-32x+178=0 are x=8+5i,x=8-5i

Find the zeros of 2x^2-32x+178=0

The zeros of 2x^2-32x+178=0 are x=8+5i,x=8-5i

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2 x^{2} - 32 x + 178 = 0

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

Solution steps

2 x^{2} - 32 x + 178 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 2 \cdot 178}{2 \cdot 2}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} = 9 - x^{2}

2 x^{2} - 6 x - 16 = 0

3 x^{2} - 5 x = 1

2 x^{2} + 24 x - 3

z^{2} + 18 = 11 z

What are the solutions to the equation 2x^2-32x+178=0 ?
The solutions to the equation 2x^2-32x+178=0 are x=8+5i,x=8-5i
Find the zeros of 2x^2-32x+178=0
The zeros of 2x^2-32x+178=0 are x=8+5i,x=8-5i