What are the solutions to the equation 3r^2+27r=12 ?

The solutions to the equation 3r^2+27r=12 are r=(-9+sqrt(97))/2 ,r=-(9+sqrt(97))/2

Find the zeros of 3r^2+27r=12

The zeros of 3r^2+27r=12 are r=(-9+sqrt(97))/2 ,r=-(9+sqrt(97))/2

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3 r^{2} + 27 r = 12

r = \frac{- 9 + 97}{2}, r = - \frac{9 + 97}{2}

Decimal Notation

r = 0.42442 \dots, r = - 9.42442 \dots

Solution steps

3 r^{2} + 27 r = 12

Move

12

to the left side

3 r^{2} + 27 r - 12 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

2 7^{2} - 4 \cdot 3 (- 12) = 397

r_{1, 2} = \frac{- 27 \pm 3 97}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- 27 + 3 97}{2 \cdot 3}, r_{2} = \frac{- 27 - 3 97}{2 \cdot 3}

r = \frac{- 27 + 3 97}{2 \cdot 3} : \frac{- 9 + 97}{2}

r = \frac{- 27 - 3 97}{2 \cdot 3} : - \frac{9 + 97}{2}

The solutions to the quadratic equation are:

r = \frac{- 9 + 97}{2}, r = - \frac{9 + 97}{2}

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

18 x^{2} - 9 x - 5 = 0

4 n^{2} + 8 n + 4 = 0

(3 x + 2) (3 x - 2) = 0

4 x^{2} - 13 x + 2 = 0

What are the solutions to the equation 3r^2+27r=12 ?
The solutions to the equation 3r^2+27r=12 are r=(-9+sqrt(97))/2 ,r=-(9+sqrt(97))/2
Find the zeros of 3r^2+27r=12
The zeros of 3r^2+27r=12 are r=(-9+sqrt(97))/2 ,r=-(9+sqrt(97))/2