What are the solutions to the equation 2x^2-3x+4=0 ?

The solutions to the equation 2x^2-3x+4=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

Find the zeros of 2x^2-3x+4=0

The zeros of 2x^2-3x+4=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

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2 x^{2} - 3 x + 4 = 0

x = \frac{3}{4} + i \frac{23}{4}, x = \frac{3}{4} - i \frac{23}{4}

Solution steps

2 x^{2} - 3 x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 4 : 23 i

x_{1, 2} = \frac{- ( - 3 ) \pm 23 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 23 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 23 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 23 i}{2 \cdot 2} : \frac{3}{4} + i \frac{23}{4}

x = \frac{- ( - 3 ) - 23 i}{2 \cdot 2} : \frac{3}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{23}{4}, x = \frac{3}{4} - i \frac{23}{4}

\frac{x}{x + 3} \leq \frac{1}{x - 1}

3 \cdot 8

(2 x + 1)^{2} - 9

6 - 8 (5 - x) = 7 x

- 2 \leq \frac{- 2 x + 1}{- 3} \leq 4

What are the solutions to the equation 2x^2-3x+4=0 ?
The solutions to the equation 2x^2-3x+4=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4
Find the zeros of 2x^2-3x+4=0
The zeros of 2x^2-3x+4=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4