What are the solutions to the equation 3q^2-18q+24=0 ?

The solutions to the equation 3q^2-18q+24=0 are q=4,q=2

Find the zeros of 3q^2-18q+24=0

The zeros of 3q^2-18q+24=0 are q=4,q=2

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3 q^{2} - 18 q + 24 = 0

q = 4, q = 2

Solution steps

3 q^{2} - 18 q + 24 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 3 \cdot 24}{2 \cdot 3}

(- 18)^{2} - 4 \cdot 3 \cdot 24 = 6

q_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 3}, q_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 3}

q = \frac{- ( - 18 ) + 6}{2 \cdot 3} : 4

q = \frac{- ( - 18 ) - 6}{2 \cdot 3} : 2

The solutions to the quadratic equation are:

q = 4, q = 2

0 = 1 - 0^{2} - z^{2}

5 x^{2} - 7 x = - 4 + 3 x^{2}

- 11 a^{2} + 32 a + 256 = 0

- 7 x^{2} - 21 x - 5

3 x + 5 = 4 x^{2}

What are the solutions to the equation 3q^2-18q+24=0 ?
The solutions to the equation 3q^2-18q+24=0 are q=4,q=2
Find the zeros of 3q^2-18q+24=0
The zeros of 3q^2-18q+24=0 are q=4,q=2