What are the solutions to the equation 3q^2+q-14=0 ?

The solutions to the equation 3q^2+q-14=0 are q=2,q=-7/3

Find the zeros of 3q^2+q-14=0

The zeros of 3q^2+q-14=0 are q=2,q=-7/3

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3 q^{2} + q - 14 = 0

q = 2, q = - \frac{7}{3}

Solution steps

3 q^{2} + q - 14 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 14 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 14) = 13

q_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- 1 + 13}{2 \cdot 3}, q_{2} = \frac{- 1 - 13}{2 \cdot 3}

q = \frac{- 1 + 13}{2 \cdot 3} : 2

q = \frac{- 1 - 13}{2 \cdot 3} : - \frac{7}{3}

The solutions to the quadratic equation are:

q = 2, q = - \frac{7}{3}

x^{4} + 3 x^{3} - 5 x^{2} - 1 = 0

\frac{k}{6} = \frac{4}{3}

(- 4 x^{3} y^{5})^{4}

∣ 3 x - 5 ∣ + 2 = 4 x

- 2 m + 5 = - 2 m + 5

What are the solutions to the equation 3q^2+q-14=0 ?
The solutions to the equation 3q^2+q-14=0 are q=2,q=-7/3
Find the zeros of 3q^2+q-14=0
The zeros of 3q^2+q-14=0 are q=2,q=-7/3