What are the solutions to the equation q^2-7q+3=-2q ?

The solutions to the equation q^2-7q+3=-2q are q=(5+sqrt(13))/2 ,q=(5-sqrt(13))/2

Find the zeros of q^2-7q+3=-2q

The zeros of q^2-7q+3=-2q are q=(5+sqrt(13))/2 ,q=(5-sqrt(13))/2

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q^{2} - 7 q + 3 = - 2 q

q = \frac{5 + 13}{2}, q = \frac{5 - 13}{2}

Decimal Notation

q = 4.30277 \dots, q = 0.69722 \dots

Solution steps

q^{2} - 7 q + 3 = - 2 q

Move

2 q

to the left side

q^{2} - 5 q + 3 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 3 = 13

q_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 1}, q_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 1}

q = \frac{- ( - 5 ) + 13}{2 \cdot 1} : \frac{5 + 13}{2}

q = \frac{- ( - 5 ) - 13}{2 \cdot 1} : \frac{5 - 13}{2}

The solutions to the quadratic equation are:

q = \frac{5 + 13}{2}, q = \frac{5 - 13}{2}

x^{2} + 9 x + c

x^{2} + 12 x + 27 = 0

2 + x = 6 x^{2}

13 x^{2} + 78 x = 0

x, x^{2} + 18 x + 81 = 25

What are the solutions to the equation q^2-7q+3=-2q ?
The solutions to the equation q^2-7q+3=-2q are q=(5+sqrt(13))/2 ,q=(5-sqrt(13))/2
Find the zeros of q^2-7q+3=-2q
The zeros of q^2-7q+3=-2q are q=(5+sqrt(13))/2 ,q=(5-sqrt(13))/2