What are the solutions to the equation 3Q^2+5Q+7=-Q+16 ?

The solutions to the equation 3Q^2+5Q+7=-Q+16 are Q=1,Q=-3

Find the zeros of 3Q^2+5Q+7=-Q+16

The zeros of 3Q^2+5Q+7=-Q+16 are Q=1,Q=-3

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3 Q^{2} + 5 Q + 7 = - Q + 16

Q = 1, Q = - 3

Solution steps

3 Q^{2} + 5 Q + 7 = - Q + 16

Move

16

to the left side

3 Q^{2} + 5 Q - 9 = - Q

Move

Q

to the left side

3 Q^{2} + 6 Q - 9 = 0

Solve with the quadratic formula

Q_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 9) = 12

Q_{1, 2} = \frac{- 6 \pm 12}{2 \cdot 3}

Separate the solutions

Q_{1} = \frac{- 6 + 12}{2 \cdot 3}, Q_{2} = \frac{- 6 - 12}{2 \cdot 3}

Q = \frac{- 6 + 12}{2 \cdot 3} : 1

Q = \frac{- 6 - 12}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

Q = 1, Q = - 3

z, x^{2} + y^{2} + z^{2} = 16

4 m^{2} + 6 = 4 m + 1

x^{2} + 21 x + 38 = 0

m^{2} - 12 m + 11 = 0

(12 m - 4 m^{2})^{2} - 4 (1 + m^{2}) (4 m^{2} - 24 m + 20) = 0

What are the solutions to the equation 3Q^2+5Q+7=-Q+16 ?
The solutions to the equation 3Q^2+5Q+7=-Q+16 are Q=1,Q=-3
Find the zeros of 3Q^2+5Q+7=-Q+16
The zeros of 3Q^2+5Q+7=-Q+16 are Q=1,Q=-3