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

The solutions to the equation 4a^2-2a+3=0 are a= 1/4+i(sqrt(11))/4 ,a= 1/4-i(sqrt(11))/4

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

The zeros of 4a^2-2a+3=0 are a= 1/4+i(sqrt(11))/4 ,a= 1/4-i(sqrt(11))/4

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

a = \frac{1}{4} + i \frac{11}{4}, a = \frac{1}{4} - i \frac{11}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 4 \cdot 3 : 211 i

a_{1, 2} = \frac{- ( - 2 ) \pm 2 11 i}{2 \cdot 4}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 2 11 i}{2 \cdot 4}, a_{2} = \frac{- ( - 2 ) - 2 11 i}{2 \cdot 4}

a = \frac{- ( - 2 ) + 2 11 i}{2 \cdot 4} : \frac{1}{4} + i \frac{11}{4}

a = \frac{- ( - 2 ) - 2 11 i}{2 \cdot 4} : \frac{1}{4} - i \frac{11}{4}

The solutions to the quadratic equation are:

a = \frac{1}{4} + i \frac{11}{4}, a = \frac{1}{4} - i \frac{11}{4}

3 = y^{2} + 2 y

\frac{n ( n - 3 )}{2} - n = 25

(5 p + 1)^{2} = 12

81 x^{2} = 169

r^{2} - 3 r = 0

What are the solutions to the equation 4a^2-2a+3=0 ?
The solutions to the equation 4a^2-2a+3=0 are a= 1/4+i(sqrt(11))/4 ,a= 1/4-i(sqrt(11))/4
Find the zeros of 4a^2-2a+3=0
The zeros of 4a^2-2a+3=0 are a= 1/4+i(sqrt(11))/4 ,a= 1/4-i(sqrt(11))/4