# 14x-17+3x^2=19+11x equation

The teacher will be very surprised to see your correct solution 😉

v

#### Numerical solution:

Do search numerical solution at [, ]

### The solution

You have entered [src]
               2
14*x - 17 + 3*x  = 19 + 11*x
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
Detail solution
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
to
$$\left(- 11 x - 19\right) + \left(3 x^{2} + 14 x - 17\right) = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3$$
$$b = 3$$
$$c = -36$$
, then
$$D = b^2 - 4\ a\ c =$$
$$3^{2} - 3 \cdot 4 \left(-36\right) = 441$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 3$$
Simplify
$$x_{2} = -4$$
Simplify
Vieta's Theorem
rewrite the equation
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
of
$$a x^{2} + b x + c = 0$$
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + x - 12 = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 1$$
$$q = \frac{c}{a}$$
$$q = -12$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -1$$
$$x_{1} x_{2} = -12$$
The graph
Sum and product of roots [src]
sum
-4 + 3
$$\left(-4\right) + \left(3\right)$$
=
-1
$$-1$$
product
-4 * 3
$$\left(-4\right) * \left(3\right)$$
=
-12
$$-12$$
Rapid solution [src]
x_1 = -4
$$x_{1} = -4$$
x_2 = 3
$$x_{2} = 3$$
x1 = 3.0
x2 = -4.0
x2 = -4.0 