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(x^2-3x-(4+3x))=0 equation
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The solution
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[src]
2 x - 3*x + -4 - 3*x = 0
$$\left(- 3 x - 4\right) + \left(x^{2} - 3 x\right) = 0$$
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -6$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 3 + \sqrt{13}$$
$$x_{2} = 3 - \sqrt{13}$$
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -6$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(-6)^2 - 4 * (1) * (-4) = 52
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3 + \sqrt{13}$$
$$x_{2} = 3 - \sqrt{13}$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = -4$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 6$$
$$x_{1} x_{2} = -4$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = -4$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 6$$
$$x_{1} x_{2} = -4$$
Rapid solution
[src]
____ x1 = 3 - \/ 13
$$x_{1} = 3 - \sqrt{13}$$
____ x2 = 3 + \/ 13
$$x_{2} = 3 + \sqrt{13}$$
x2 = 3 + sqrt(13)
Sum and product of roots
[src]
sum
____ ____ 3 - \/ 13 + 3 + \/ 13
$$\left(3 - \sqrt{13}\right) + \left(3 + \sqrt{13}\right)$$
=
6
$$6$$
product
/ ____\ / ____\ \3 - \/ 13 /*\3 + \/ 13 /
$$\left(3 - \sqrt{13}\right) \left(3 + \sqrt{13}\right)$$
=
-4
$$-4$$
-4