What are the solutions to the equation 4x^2-5x+6=0 ?

The solutions to the equation 4x^2-5x+6=0 are x= 5/8+i(sqrt(71))/8 ,x= 5/8-i(sqrt(71))/8

Find the zeros of 4x^2-5x+6=0

The zeros of 4x^2-5x+6=0 are x= 5/8+i(sqrt(71))/8 ,x= 5/8-i(sqrt(71))/8

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4 x^{2} - 5 x + 6 = 0

x = \frac{5}{8} + i \frac{71}{8}, x = \frac{5}{8} - i \frac{71}{8}

Solution steps

4 x^{2} - 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 6}{2 \cdot 4}

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 6 : 71 i

x_{1, 2} = \frac{- ( - 5 ) \pm 71 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 71 i}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 71 i}{2 \cdot 4}

x = \frac{- ( - 5 ) + 71 i}{2 \cdot 4} : \frac{5}{8} + i \frac{71}{8}

x = \frac{- ( - 5 ) - 71 i}{2 \cdot 4} : \frac{5}{8} - i \frac{71}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} + i \frac{71}{8}, x = \frac{5}{8} - i \frac{71}{8}

3 x + 3 = 12

3 x + \frac{4}{5} = 7 - 2 x

2 x + 3 (x - 10) = 45

- 4 (x + 2) + 2 = 2 (x + 6)

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

What are the solutions to the equation 4x^2-5x+6=0 ?
The solutions to the equation 4x^2-5x+6=0 are x= 5/8+i(sqrt(71))/8 ,x= 5/8-i(sqrt(71))/8
Find the zeros of 4x^2-5x+6=0
The zeros of 4x^2-5x+6=0 are x= 5/8+i(sqrt(71))/8 ,x= 5/8-i(sqrt(71))/8