What are the solutions to the equation 9q^2-11q+2=3q^2 ?

The solutions to the equation 9q^2-11q+2=3q^2 are q=(11+sqrt(73))/(12),q=(11-sqrt(73))/(12)

Find the zeros of 9q^2-11q+2=3q^2

The zeros of 9q^2-11q+2=3q^2 are q=(11+sqrt(73))/(12),q=(11-sqrt(73))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9 q^{2} - 11 q + 2 = 3 q^{2}

q = \frac{11 + 73}{12}, q = \frac{11 - 73}{12}

Decimal Notation

q = 1.62866 \dots, q = 0.20466 \dots

Solution steps

9 q^{2} - 11 q + 2 = 3 q^{2}

Move

3 q^{2}

to the left side

6 q^{2} - 11 q + 2 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

(- 11)^{2} - 4 \cdot 6 \cdot 2 = 73

q_{1, 2} = \frac{- ( - 11 ) \pm 73}{2 \cdot 6}

Separate the solutions

q_{1} = \frac{- ( - 11 ) + 73}{2 \cdot 6}, q_{2} = \frac{- ( - 11 ) - 73}{2 \cdot 6}

q = \frac{- ( - 11 ) + 73}{2 \cdot 6} : \frac{11 + 73}{12}

q = \frac{- ( - 11 ) - 73}{2 \cdot 6} : \frac{11 - 73}{12}

The solutions to the quadratic equation are:

q = \frac{11 + 73}{12}, q = \frac{11 - 73}{12}

7 x^{2} - 9 x + 3 = 0

(x + 2)^{2} + 3 = 0

x^{2} + 8 x = - 24

x^{2} + 8 x = - 10

y, x^{2} + y^{2} - 4 x + 6 y + 13 = 0

What are the solutions to the equation 9q^2-11q+2=3q^2 ?
The solutions to the equation 9q^2-11q+2=3q^2 are q=(11+sqrt(73))/(12),q=(11-sqrt(73))/(12)
Find the zeros of 9q^2-11q+2=3q^2
The zeros of 9q^2-11q+2=3q^2 are q=(11+sqrt(73))/(12),q=(11-sqrt(73))/(12)