What are the solutions to the equation 2x^2-4x+7=0 ?

The solutions to the equation 2x^2-4x+7=0 are x=1+isqrt(5/2),x=1-isqrt(5/2)

Find the zeros of 2x^2-4x+7=0

The zeros of 2x^2-4x+7=0 are x=1+isqrt(5/2),x=1-isqrt(5/2)

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2 x^{2} - 4 x + 7 = 0

x = 1 + i \frac{5}{2}, x = 1 - i \frac{5}{2}

Solution steps

2 x^{2} - 4 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 7 : 210 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 10 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 10 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 10 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 10 i}{2 \cdot 2} : 1 + i \frac{5}{2}

x = \frac{- ( - 4 ) - 2 10 i}{2 \cdot 2} : 1 - i \frac{5}{2}

The solutions to the quadratic equation are:

x = 1 + i \frac{5}{2}, x = 1 - i \frac{5}{2}

∣ x ∣ < 12

(\frac{y}{2})^{- 10}

20 x^{2} + 22 x - 12

- 3 - 3 x \geq - 7

∣ 3 x - 2 ∣ > 13

What are the solutions to the equation 2x^2-4x+7=0 ?
The solutions to the equation 2x^2-4x+7=0 are x=1+isqrt(5/2),x=1-isqrt(5/2)
Find the zeros of 2x^2-4x+7=0
The zeros of 2x^2-4x+7=0 are x=1+isqrt(5/2),x=1-isqrt(5/2)