What are the solutions to the equation 6q^2-18q+12=0 ?

The solutions to the equation 6q^2-18q+12=0 are q=2,q=1

Find the zeros of 6q^2-18q+12=0

The zeros of 6q^2-18q+12=0 are q=2,q=1

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6 q^{2} - 18 q + 12 = 0

q = 2, q = 1

Solution steps

6 q^{2} - 18 q + 12 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

(- 18)^{2} - 4 \cdot 6 \cdot 12 = 6

q_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 6}

Separate the solutions

q_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 6}, q_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 6}

q = \frac{- ( - 18 ) + 6}{2 \cdot 6} : 2

q = \frac{- ( - 18 ) - 6}{2 \cdot 6} : 1

The solutions to the quadratic equation are:

q = 2, q = 1

x (x + 1) = 342

9 x^{2} = - 9 x

x^{2} - 10 x + 63 = 6 x

5 x^{2} + 5 x - 3 = - 8 x - 9

x^{2} - 13 x + 40 = 0

What are the solutions to the equation 6q^2-18q+12=0 ?
The solutions to the equation 6q^2-18q+12=0 are q=2,q=1
Find the zeros of 6q^2-18q+12=0
The zeros of 6q^2-18q+12=0 are q=2,q=1