What are the solutions to the equation 7q^2+6q+3=6q^2 ?

The solutions to the equation 7q^2+6q+3=6q^2 are q=-3+sqrt(6),q=-3-sqrt(6)

Find the zeros of 7q^2+6q+3=6q^2

The zeros of 7q^2+6q+3=6q^2 are q=-3+sqrt(6),q=-3-sqrt(6)

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7 q^{2} + 6 q + 3 = 6 q^{2}

q = - 3 + 6, q = - 3 - 6

Decimal Notation

q = - 0.55051 \dots, q = - 5.44948 \dots

Solution steps

7 q^{2} + 6 q + 3 = 6 q^{2}

Move

6 q^{2}

to the left side

q^{2} + 6 q + 3 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 3 = 26

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

Separate the solutions

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

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

q = \frac{- 6 - 2 6}{2 \cdot 1} : - 3 - 6

The solutions to the quadratic equation are:

q = - 3 + 6, q = - 3 - 6

x^{2} + 9 x + 35 = 0

x^{2} + 2 x = 28

9 p^{2} + 6 p + 1 = 12

x^{2} = 9 x + 2

(14 x + 21) (1.8 - 0.3 x) = 0

What are the solutions to the equation 7q^2+6q+3=6q^2 ?
The solutions to the equation 7q^2+6q+3=6q^2 are q=-3+sqrt(6),q=-3-sqrt(6)
Find the zeros of 7q^2+6q+3=6q^2
The zeros of 7q^2+6q+3=6q^2 are q=-3+sqrt(6),q=-3-sqrt(6)