What are the solutions to the equation 2v^2-3v+6=0 ?

The solutions to the equation 2v^2-3v+6=0 are v= 3/4+i(sqrt(39))/4 ,v= 3/4-i(sqrt(39))/4

Find the zeros of 2v^2-3v+6=0

The zeros of 2v^2-3v+6=0 are v= 3/4+i(sqrt(39))/4 ,v= 3/4-i(sqrt(39))/4

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2 v^{2} - 3 v + 6 = 0

v = \frac{3}{4} + i \frac{39}{4}, v = \frac{3}{4} - i \frac{39}{4}

Solution steps

2 v^{2} - 3 v + 6 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 6 : 39 i

v_{1, 2} = \frac{- ( - 3 ) \pm 39 i}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 39 i}{2 \cdot 2}, v_{2} = \frac{- ( - 3 ) - 39 i}{2 \cdot 2}

v = \frac{- ( - 3 ) + 39 i}{2 \cdot 2} : \frac{3}{4} + i \frac{39}{4}

v = \frac{- ( - 3 ) - 39 i}{2 \cdot 2} : \frac{3}{4} - i \frac{39}{4}

The solutions to the quadratic equation are:

v = \frac{3}{4} + i \frac{39}{4}, v = \frac{3}{4} - i \frac{39}{4}

- 2 m^{2} - 28 m - 98 = 0

10 x^{2} + 6 x - 4 = 0

x^{2} + 11 x = - 24

x^{2} - (8 \cdot x) = 9

x^{2} - 16 = 0

What are the solutions to the equation 2v^2-3v+6=0 ?
The solutions to the equation 2v^2-3v+6=0 are v= 3/4+i(sqrt(39))/4 ,v= 3/4-i(sqrt(39))/4
Find the zeros of 2v^2-3v+6=0
The zeros of 2v^2-3v+6=0 are v= 3/4+i(sqrt(39))/4 ,v= 3/4-i(sqrt(39))/4