What are the solutions to the equation 18x^2-54x+36=0 ?

The solutions to the equation 18x^2-54x+36=0 are x=2,x=1

Find the zeros of 18x^2-54x+36=0

The zeros of 18x^2-54x+36=0 are x=2,x=1

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18 x^{2} - 54 x + 36 = 0

x = 2, x = 1

Solution steps

18 x^{2} - 54 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 54 ) \pm ( - 54 ) ^{2} - 4 \cdot 18 \cdot 36}{2 \cdot 18}

(- 54)^{2} - 4 \cdot 18 \cdot 36 = 18

x_{1, 2} = \frac{- ( - 54 ) \pm 18}{2 \cdot 18}

Separate the solutions

x_{1} = \frac{- ( - 54 ) + 18}{2 \cdot 18}, x_{2} = \frac{- ( - 54 ) - 18}{2 \cdot 18}

x = \frac{- ( - 54 ) + 18}{2 \cdot 18} : 2

x = \frac{- ( - 54 ) - 18}{2 \cdot 18} : 1

The solutions to the quadratic equation are:

x = 2, x = 1

(25 x)^{\frac{1}{2}}

3 x + 9 = x + 1

\frac{x}{5} + (1 - x) = - 27

4 x^{2} - 20 x = - 13

\frac{x + 1}{4} = \frac{1}{6} + \frac{2 - x}{3}

What are the solutions to the equation 18x^2-54x+36=0 ?
The solutions to the equation 18x^2-54x+36=0 are x=2,x=1
Find the zeros of 18x^2-54x+36=0
The zeros of 18x^2-54x+36=0 are x=2,x=1