What are the solutions to the equation 6x^2-9x+12=0 ?

The solutions to the equation 6x^2-9x+12=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

Find the zeros of 6x^2-9x+12=0

The zeros of 6x^2-9x+12=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

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6 x^{2} - 9 x + 12 = 0

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

Solution steps

6 x^{2} - 9 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

Simplify

(- 9)^{2} - 4 \cdot 6 \cdot 12 : 323 i

x_{1, 2} = \frac{- ( - 9 ) \pm 3 23 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3 23 i}{2 \cdot 6}, x_{2} = \frac{- ( - 9 ) - 3 23 i}{2 \cdot 6}

x = \frac{- ( - 9 ) + 3 23 i}{2 \cdot 6} : \frac{3}{4} + i \frac{23}{4}

x = \frac{- ( - 9 ) - 3 23 i}{2 \cdot 6} : \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}

x^{2} + 5 x - (a^{2} + a - 6) = 0

y, x^{3} - x y^{2} + 3 y^{2} + 2 = 0

\frac{2 x ^{2} - ( k + 3 ) x - k + 5}{4} = 0

(x + 3 a) (x - 4 b) = 0

x, y^{11} + y^{1} + y = x^{2}

What are the solutions to the equation 6x^2-9x+12=0 ?
The solutions to the equation 6x^2-9x+12=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4
Find the zeros of 6x^2-9x+12=0
The zeros of 6x^2-9x+12=0 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4