What are the solutions to the equation 3x^2-36x+81=0 ?

The solutions to the equation 3x^2-36x+81=0 are x=9,x=3

Find the zeros of 3x^2-36x+81=0

The zeros of 3x^2-36x+81=0 are x=9,x=3

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3 x^{2} - 36 x + 81 = 0

x = 9, x = 3

Solution steps

3 x^{2} - 36 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 3 \cdot 81}{2 \cdot 3}

(- 36)^{2} - 4 \cdot 3 \cdot 81 = 18

x_{1, 2} = \frac{- ( - 36 ) \pm 18}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 18}{2 \cdot 3}, x_{2} = \frac{- ( - 36 ) - 18}{2 \cdot 3}

x = \frac{- ( - 36 ) + 18}{2 \cdot 3} : 9

x = \frac{- ( - 36 ) - 18}{2 \cdot 3} : 3

The solutions to the quadratic equation are:

x = 9, x = 3

e^{2^{2}}

4 (0) + y = 5

7 x^{2} + 8 x + 1 = 0

2^{4}

3 (y - 4) = - 2 (y + 10)

What are the solutions to the equation 3x^2-36x+81=0 ?
The solutions to the equation 3x^2-36x+81=0 are x=9,x=3
Find the zeros of 3x^2-36x+81=0
The zeros of 3x^2-36x+81=0 are x=9,x=3