What are the solutions to the equation (q(q+7))/7-q=4-q^2 ?

The solutions to the equation (q(q+7))/7-q=4-q^2 are q=sqrt(7/2),q=-sqrt(7/2)

Find the zeros of (q(q+7))/7-q=4-q^2

The zeros of (q(q+7))/7-q=4-q^2 are q=sqrt(7/2),q=-sqrt(7/2)

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\frac{q ( q + 7 )}{7} - q = 4 - q^{2}

q = \frac{7}{2}, q = - \frac{7}{2}

Decimal Notation

q = 1.87082 \dots, q = - 1.87082 \dots

Solution steps

\frac{q ( q + 7 )}{7} - q = 4 - q^{2}

Expand

\frac{q ( q + 7 )}{7} - q : \frac{q ^{2}}{7}

\frac{q ^{2}}{7} = 4 - q^{2}

Move

q^{2}

to the left side

\frac{8 q ^{2}}{7} = 4

Multiply both sides by

7

8 q^{2} = 28

Divide both sides by

8

q^{2} = \frac{7}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

q = \frac{7}{2}, q = - \frac{7}{2}

3 y^{2} + 9 y + 15

a x^{2} - (a + 2) x - 1 = 0

x, x^{2} y + x - y + 1 = 0

z^{2} - 11 z = 0

x^{2} = (531 mod 2021)

What are the solutions to the equation (q(q+7))/7-q=4-q^2 ?
The solutions to the equation (q(q+7))/7-q=4-q^2 are q=sqrt(7/2),q=-sqrt(7/2)
Find the zeros of (q(q+7))/7-q=4-q^2
The zeros of (q(q+7))/7-q=4-q^2 are q=sqrt(7/2),q=-sqrt(7/2)