3y²+7y-6 equation
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The solution
Detail solution
This equation is of the form
a*y^2 + b*y + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = 7$$
$$c = -6$$
, then
D = b^2 - 4 * a * c =
(7)^2 - 4 * (3) * (-6) = 121
Because D > 0, then the equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = \frac{2}{3}$$
$$y_{2} = -3$$
Vieta's Theorem
rewrite the equation
$$\left(3 y^{2} + 7 y\right) - 6 = 0$$
of
$$a y^{2} + b y + c = 0$$
as reduced quadratic equation
$$y^{2} + \frac{b y}{a} + \frac{c}{a} = 0$$
$$y^{2} + \frac{7 y}{3} - 2 = 0$$
$$p y + q + y^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{7}{3}$$
$$q = \frac{c}{a}$$
$$q = -2$$
Vieta Formulas
$$y_{1} + y_{2} = - p$$
$$y_{1} y_{2} = q$$
$$y_{1} + y_{2} = - \frac{7}{3}$$
$$y_{1} y_{2} = -2$$
$$y_{1} = -3$$
$$y_{2} = \frac{2}{3}$$
Sum and product of roots
[src]
$$-3 + \frac{2}{3}$$
$$- \frac{7}{3}$$
$$- 2$$
$$-2$$