What are the solutions to the equation 2x^2-7x+39=0 ?

The solutions to the equation 2x^2-7x+39=0 are x= 7/4+i(sqrt(263))/4 ,x= 7/4-i(sqrt(263))/4

Find the zeros of 2x^2-7x+39=0

The zeros of 2x^2-7x+39=0 are x= 7/4+i(sqrt(263))/4 ,x= 7/4-i(sqrt(263))/4

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2 x^{2} - 7 x + 39 = 0

x = \frac{7}{4} + i \frac{263}{4}, x = \frac{7}{4} - i \frac{263}{4}

Solution steps

2 x^{2} - 7 x + 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 39}{2 \cdot 2}

Simplify

(- 7)^{2} - 4 \cdot 2 \cdot 39 : 263 i

x_{1, 2} = \frac{- ( - 7 ) \pm 263 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 263 i}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 263 i}{2 \cdot 2}

x = \frac{- ( - 7 ) + 263 i}{2 \cdot 2} : \frac{7}{4} + i \frac{263}{4}

x = \frac{- ( - 7 ) - 263 i}{2 \cdot 2} : \frac{7}{4} - i \frac{263}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} + i \frac{263}{4}, x = \frac{7}{4} - i \frac{263}{4}

28 - x^{3} = (4 - x)^{3}

4 (\frac{1}{3})^{2} + y^{2} = 4

1 = x^{2} - 3

1 = x^{2} - 4

5 a^{2} - 12 a + 3 = 0

What are the solutions to the equation 2x^2-7x+39=0 ?
The solutions to the equation 2x^2-7x+39=0 are x= 7/4+i(sqrt(263))/4 ,x= 7/4-i(sqrt(263))/4
Find the zeros of 2x^2-7x+39=0
The zeros of 2x^2-7x+39=0 are x= 7/4+i(sqrt(263))/4 ,x= 7/4-i(sqrt(263))/4