What are the solutions to the equation q^2-3q-11=-6q ?

The solutions to the equation q^2-3q-11=-6q are q=(-3+sqrt(53))/2 ,q=(-3-sqrt(53))/2

Find the zeros of q^2-3q-11=-6q

The zeros of q^2-3q-11=-6q are q=(-3+sqrt(53))/2 ,q=(-3-sqrt(53))/2

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q^{2} - 3 q - 11 = - 6 q

q = \frac{- 3 + 53}{2}, q = \frac{- 3 - 53}{2}

Decimal Notation

q = 2.14005 \dots, q = - 5.14005 \dots

Solution steps

q^{2} - 3 q - 11 = - 6 q

Move

6 q

to the left side

q^{2} + 3 q - 11 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 11) = 53

q_{1, 2} = \frac{- 3 \pm 53}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- 3 + 53}{2 \cdot 1}, q_{2} = \frac{- 3 - 53}{2 \cdot 1}

q = \frac{- 3 + 53}{2 \cdot 1} : \frac{- 3 + 53}{2}

q = \frac{- 3 - 53}{2 \cdot 1} : \frac{- 3 - 53}{2}

The solutions to the quadratic equation are:

q = \frac{- 3 + 53}{2}, q = \frac{- 3 - 53}{2}

(36 - r)^{2} + (18 - r)^{2} = r^{2}

(62)^{2} = x^{2} + x^{2}

15 + 7 k = 2 k^{2}

2 x^{2} + 12 x + 2 = 0

4 p^{2} - 5 p - 6 = - 2

What are the solutions to the equation q^2-3q-11=-6q ?
The solutions to the equation q^2-3q-11=-6q are q=(-3+sqrt(53))/2 ,q=(-3-sqrt(53))/2
Find the zeros of q^2-3q-11=-6q
The zeros of q^2-3q-11=-6q are q=(-3+sqrt(53))/2 ,q=(-3-sqrt(53))/2