What are the solutions to the equation 121x^2-132x+36=0 ?

The solutions to the equation 121x^2-132x+36=0 are x= 6/11

Find the zeros of 121x^2-132x+36=0

The zeros of 121x^2-132x+36=0 are x= 6/11

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121 x^{2} - 132 x + 36 = 0

x = \frac{6}{11}

Decimal Notation

x = 0.54545 \dots

Solution steps

121 x^{2} - 132 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 132 ) \pm ( - 132 ) ^{2} - 4 \cdot 121 \cdot 36}{2 \cdot 121}

(- 132)^{2} - 4 \cdot 121 \cdot 36 = 0

x_{1, 2} = \frac{- ( - 132 ) \pm 0}{2 \cdot 121}

x = \frac{- ( - 132 )}{2 \cdot 121}

\frac{- ( - 132 )}{2 \cdot 121} = \frac{6}{11}

x = \frac{6}{11}

The solution to the quadratic equation is:

x = \frac{6}{11}

(n + 10) + 2 (n + 2) + (n + 4)^{2} = 1984

3 x^{2} - 8 x - 14 = 0

- y (5 - y) = 0

6 x^{2} = - 5 + 11 x

p^{2} + 7 p + 2 = 0

What are the solutions to the equation 121x^2-132x+36=0 ?
The solutions to the equation 121x^2-132x+36=0 are x= 6/11
Find the zeros of 121x^2-132x+36=0
The zeros of 121x^2-132x+36=0 are x= 6/11