What are the solutions to the equation q(q+7)=5q^2-3 ?

The solutions to the equation q(q+7)=5q^2-3 are q=-(-7+sqrt(97))/8 ,q=(7+sqrt(97))/8

Find the zeros of q(q+7)=5q^2-3

The zeros of q(q+7)=5q^2-3 are q=-(-7+sqrt(97))/8 ,q=(7+sqrt(97))/8

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q (q + 7) = 5 q^{2} - 3

q = - \frac{- 7 + 97}{8}, q = \frac{7 + 97}{8}

Decimal Notation

q = - 0.35610 \dots, q = 2.10610 \dots

Solution steps

q (q + 7) = 5 q^{2} - 3

Expand

q (q + 7) : q^{2} + 7 q

q^{2} + 7 q = 5 q^{2} - 3

Move

3

to the left side

q^{2} + 7 q + 3 = 5 q^{2}

Move

5 q^{2}

to the left side

- 4 q^{2} + 7 q + 3 = 0

Solve with the quadratic formula

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

7^{2} - 4 (- 4) \cdot 3 = 97

q_{1, 2} = \frac{- 7 \pm 97}{2 ( - 4 )}

Separate the solutions

q_{1} = \frac{- 7 + 97}{2 ( - 4 )}, q_{2} = \frac{- 7 - 97}{2 ( - 4 )}

q = \frac{- 7 + 97}{2 ( - 4 )} : - \frac{- 7 + 97}{8}

q = \frac{- 7 - 97}{2 ( - 4 )} : \frac{7 + 97}{8}

The solutions to the quadratic equation are:

q = - \frac{- 7 + 97}{8}, q = \frac{7 + 97}{8}

3 x^{2} - x^{2} + 7 x + 3 = 4

3 x^{2} - 6 x - 22 = 0

(x + 8) (x - 5) = 0

- 2 x^{2} + 4 x - 9 = 0

2 x^{2} = 9 x + 18

What are the solutions to the equation q(q+7)=5q^2-3 ?
The solutions to the equation q(q+7)=5q^2-3 are q=-(-7+sqrt(97))/8 ,q=(7+sqrt(97))/8
Find the zeros of q(q+7)=5q^2-3
The zeros of q(q+7)=5q^2-3 are q=-(-7+sqrt(97))/8 ,q=(7+sqrt(97))/8