What are the solutions to the equation 9r^2+6r=1 ?

The solutions to the equation 9r^2+6r=1 are r=(-1+sqrt(2))/3 ,r=-(1+sqrt(2))/3

Find the zeros of 9r^2+6r=1

The zeros of 9r^2+6r=1 are r=(-1+sqrt(2))/3 ,r=-(1+sqrt(2))/3

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9 r^{2} + 6 r = 1

r = \frac{- 1 + 2}{3}, r = - \frac{1 + 2}{3}

Decimal Notation

r = 0.13807 \dots, r = - 0.80473 \dots

Solution steps

9 r^{2} + 6 r = 1

Move

1

to the left side

9 r^{2} + 6 r - 1 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 9 (- 1) = 62

r_{1, 2} = \frac{- 6 \pm 6 2}{2 \cdot 9}

Separate the solutions

r_{1} = \frac{- 6 + 6 2}{2 \cdot 9}, r_{2} = \frac{- 6 - 6 2}{2 \cdot 9}

r = \frac{- 6 + 6 2}{2 \cdot 9} : \frac{- 1 + 2}{3}

r = \frac{- 6 - 6 2}{2 \cdot 9} : - \frac{1 + 2}{3}

The solutions to the quadratic equation are:

r = \frac{- 1 + 2}{3}, r = - \frac{1 + 2}{3}

2 x^{2} + x = 5 x^{2} - 2 x

x^{2} - 100 x + 2499.96 = 0

3 x^{2} - 14 x - 24 = (3 x + 4)^{2}

\frac{1}{5} (x - 1)^{2} - 1 = 4

- 9 r + r^{2} + 23 = 0

What are the solutions to the equation 9r^2+6r=1 ?
The solutions to the equation 9r^2+6r=1 are r=(-1+sqrt(2))/3 ,r=-(1+sqrt(2))/3
Find the zeros of 9r^2+6r=1
The zeros of 9r^2+6r=1 are r=(-1+sqrt(2))/3 ,r=-(1+sqrt(2))/3