What are the solutions to the equation x^{(2)}-4x*3^{(1)}+27=0 ?

The solutions to the equation x^{(2)}-4x*3^{(1)}+27=0 are x=9,x=3

Find the zeros of x^{(2)}-4x*3^{(1)}+27=0

The zeros of x^{(2)}-4x*3^{(1)}+27=0 are x=9,x=3

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x^{(2)} - 4 x \cdot 3^{(1)} + 27 = 0

x = 9, x = 3

Solution steps

x^{(2)} - 4 x \cdot 3^{(1)} + 27 = 0

Expand

x^{2} - 4 x \cdot 3^{1} + 27 : x^{2} - 12 x + 27

x^{2} - 12 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 27 = 6

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

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 1}

x = \frac{- ( - 12 ) + 6}{2 \cdot 1} : 9

x = \frac{- ( - 12 ) - 6}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 9, x = 3

3 (x + 2) (x + 3) = 60

9 = 3 + 21 t - 16 t^{2}

6 w^{2} - 8 w + 5 = 0

x, 4 x y + x^{2} - 3 y^{2} = 2

4 x^{2} - 72 = 0

What are the solutions to the equation x^{(2)}-4x*3^{(1)}+27=0 ?
The solutions to the equation x^{(2)}-4x*3^{(1)}+27=0 are x=9,x=3
Find the zeros of x^{(2)}-4x*3^{(1)}+27=0
The zeros of x^{(2)}-4x*3^{(1)}+27=0 are x=9,x=3