What are the solutions to the equation 2x^2-4x+9=0 ?

The solutions to the equation 2x^2-4x+9=0 are x=1+isqrt(7/2),x=1-isqrt(7/2)

Find the zeros of 2x^2-4x+9=0

The zeros of 2x^2-4x+9=0 are x=1+isqrt(7/2),x=1-isqrt(7/2)

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2 x^{2} - 4 x + 9 = 0

x = 1 + i \frac{7}{2}, x = 1 - i \frac{7}{2}

Solution steps

2 x^{2} - 4 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 9 : 214 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 14 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 14 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 14 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 14 i}{2 \cdot 2} : 1 + i \frac{7}{2}

x = \frac{- ( - 4 ) - 2 14 i}{2 \cdot 2} : 1 - i \frac{7}{2}

The solutions to the quadratic equation are:

x = 1 + i \frac{7}{2}, x = 1 - i \frac{7}{2}

a x - a y - b y + b x

t^{2} + 8 t = - 12

- 2 + 3 (0) \geq \frac{1}{4}

\frac{5 ^{12} \cdot c ^{10} \cdot d ^{2}}{5 ^{9} \cdot c ^{9}}

2 (x + 5) - x^{2} - 5 x = 0

What are the solutions to the equation 2x^2-4x+9=0 ?
The solutions to the equation 2x^2-4x+9=0 are x=1+isqrt(7/2),x=1-isqrt(7/2)
Find the zeros of 2x^2-4x+9=0
The zeros of 2x^2-4x+9=0 are x=1+isqrt(7/2),x=1-isqrt(7/2)