What are the solutions to the equation 4x^2-5x+3=0 ?

The solutions to the equation 4x^2-5x+3=0 are x= 5/8+i(sqrt(23))/8 ,x= 5/8-i(sqrt(23))/8

Find the zeros of 4x^2-5x+3=0

The zeros of 4x^2-5x+3=0 are x= 5/8+i(sqrt(23))/8 ,x= 5/8-i(sqrt(23))/8

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4 x^{2} - 5 x + 3 = 0

x = \frac{5}{8} + i \frac{23}{8}, x = \frac{5}{8} - i \frac{23}{8}

Solution steps

4 x^{2} - 5 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 3 : 23 i

x_{1, 2} = \frac{- ( - 5 ) \pm 23 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 23 i}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 23 i}{2 \cdot 4}

x = \frac{- ( - 5 ) + 23 i}{2 \cdot 4} : \frac{5}{8} + i \frac{23}{8}

x = \frac{- ( - 5 ) - 23 i}{2 \cdot 4} : \frac{5}{8} - i \frac{23}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} + i \frac{23}{8}, x = \frac{5}{8} - i \frac{23}{8}

6 (x + 10) = 72

5 x^{2} + 1 = 6 x

- 10 s + 2 (- 2 s - 9) \leq 3 s - 1 + 7 s

\frac{1}{x + h} - \frac{1}{x}

x^{2} 10 x^{2} - 3 x - 4

What are the solutions to the equation 4x^2-5x+3=0 ?
The solutions to the equation 4x^2-5x+3=0 are x= 5/8+i(sqrt(23))/8 ,x= 5/8-i(sqrt(23))/8
Find the zeros of 4x^2-5x+3=0
The zeros of 4x^2-5x+3=0 are x= 5/8+i(sqrt(23))/8 ,x= 5/8-i(sqrt(23))/8