What are the solutions to the equation -3q^2+39q=60 ?

The solutions to the equation -3q^2+39q=60 are q=(13-sqrt(89))/2 ,q=(13+sqrt(89))/2

Find the zeros of -3q^2+39q=60

The zeros of -3q^2+39q=60 are q=(13-sqrt(89))/2 ,q=(13+sqrt(89))/2

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- 3 q^{2} + 39 q = 60

q = \frac{13 - 89}{2}, q = \frac{13 + 89}{2}

Decimal Notation

q = 1.78300 \dots, q = 11.21699 \dots

Solution steps

- 3 q^{2} + 39 q = 60

Move

60

to the left side

- 3 q^{2} + 39 q - 60 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 39 \pm 3 9 ^{2} - 4 ( - 3 ) ( - 60 )}{2 ( - 3 )}

3 9^{2} - 4 (- 3) (- 60) = 389

q_{1, 2} = \frac{- 39 \pm 3 89}{2 ( - 3 )}

Separate the solutions

q_{1} = \frac{- 39 + 3 89}{2 ( - 3 )}, q_{2} = \frac{- 39 - 3 89}{2 ( - 3 )}

q = \frac{- 39 + 3 89}{2 ( - 3 )} : \frac{13 - 89}{2}

q = \frac{- 39 - 3 89}{2 ( - 3 )} : \frac{13 + 89}{2}

The solutions to the quadratic equation are:

q = \frac{13 - 89}{2}, q = \frac{13 + 89}{2}

x^{12} - 1

lo g_{4} (49)

216 x - 12 x^{2} = 0

2 + 2 u \leq 6

x + \frac{5}{3} = \frac{2}{3}

What are the solutions to the equation -3q^2+39q=60 ?
The solutions to the equation -3q^2+39q=60 are q=(13-sqrt(89))/2 ,q=(13+sqrt(89))/2
Find the zeros of -3q^2+39q=60
The zeros of -3q^2+39q=60 are q=(13-sqrt(89))/2 ,q=(13+sqrt(89))/2