What are the solutions to the equation 36q^2-333q+81=0 ?

The solutions to the equation 36q^2-333q+81=0 are q=9,q= 1/4

Find the zeros of 36q^2-333q+81=0

The zeros of 36q^2-333q+81=0 are q=9,q= 1/4

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36 q^{2} - 333 q + 81 = 0

q = 9, q = \frac{1}{4}

Decimal

q = 9, q = 0.25

Solution steps

36 q^{2} - 333 q + 81 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 333 ) \pm ( - 333 ) ^{2} - 4 \cdot 36 \cdot 81}{2 \cdot 36}

(- 333)^{2} - 4 \cdot 36 \cdot 81 = 315

q_{1, 2} = \frac{- ( - 333 ) \pm 315}{2 \cdot 36}

Separate the solutions

q_{1} = \frac{- ( - 333 ) + 315}{2 \cdot 36}, q_{2} = \frac{- ( - 333 ) - 315}{2 \cdot 36}

q = \frac{- ( - 333 ) + 315}{2 \cdot 36} : 9

q = \frac{- ( - 333 ) - 315}{2 \cdot 36} : \frac{1}{4}

The solutions to the quadratic equation are:

q = 9, q = \frac{1}{4}

s^{2} + 2 s + 10

4 x^{2} + 69 = 28 x

x^{2} - x - 35 = - x + 1

4 x^{2} - 16 x - 33

x^{2} - 2 x - 24 = 0

What are the solutions to the equation 36q^2-333q+81=0 ?
The solutions to the equation 36q^2-333q+81=0 are q=9,q= 1/4
Find the zeros of 36q^2-333q+81=0
The zeros of 36q^2-333q+81=0 are q=9,q= 1/4