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# Solve the quadraic equation:

## $$2x^2+19x+29=0$$

Since the discriminant $$\Delta >0$$, the quadratic equation has two distinct real roots.

$$\Delta=129$$

$$\begin{cases} x_1=-\dfrac{19}{4}+\dfrac{\sqrt{129}}{4} \\ x_2=-\dfrac{19}{4}-\dfrac{\sqrt{129}}{4} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=-1.9105458270999 \\ x_2=-7.5894541729001 \end{cases}$$

Detailed Steps on Solution

## Solve the quadratic equation: $$2x² + 19x + 29 = 0$$

Given $$a =2, b=19, c=29$$,

### 1. Use the quadratic root

Use the root solution formula for a quadratic equation, the roots of the equation are given as

\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-19\pm\sqrt{19^2-4\cdot 2\cdot 29}}{2 \cdot 2}\\ & =\dfrac{-19\pm\sqrt{129}}{4}\\ & =-\dfrac{19}{4}\pm\dfrac{\sqrt{129}}{4}\\ \end{aligned}

Since the discriminat is greater than zero, we get two real roots:

That is,

$$\begin{cases} x_1 =-\dfrac{19}{4}+\dfrac{\sqrt{129}}{4} \\ x_2=-\dfrac{19}{4}-\dfrac{\sqrt{129}}{4} \end{cases}$$

### 2. Completing the square method

The idea of completing the square is to transform a quadratic equation to the form of a perfect square equals to a constant, which could be positive or minus. And then solve the equation by taking square root on both sides. For the quadratic equation,

$$2x² + 19x + 29 = 0$$

divide each term by $$2$$ to make the coefficient of the leading term $$1$$.

$$x^2+\dfrac{19}{2}x+\dfrac{29}{2}=0$$

Move the constant term $$\dfrac{29}{2}$$ to the right hand side. Then its sign becomes negative.

$$x^2+\dfrac{19}{2}x=-\dfrac{29}{2}$$

Add square of the half of $$\dfrac{19}{2}$$, the coefficient of the linear term to both sides.

$$x^2+\dfrac{19}{2}x+\Big(\dfrac{19}{4}\Big)^2=-\dfrac{29}{2}+\Big(\dfrac{19}{4}\Big)^2$$

Convert the trinomial to the form of perfect square on the left hand side. Sum on the left hand side. Then,

$$\Big(x+\dfrac{19}{4}\Big)^2=\dfrac{129}{16}$$

Taking square roots on both sides of above equation gives

$$\sqrt{\Big(x+\dfrac{19}{4}\Big)^2}=\pm\sqrt{\dfrac{129}{16}}$$

Since the left hand side is square root of a perfect square, we can get rid of radical. Then,

$$x+\dfrac{19}{4}=\pm\dfrac{\sqrt{129}}{4}$$

Move the constant $$\dfrac{19}{4}$$ to the right hand side. Then we get,

$$x_1 = -\dfrac{9}{2}$$

$$x_2 = -5$$

### 3. The vertex of the function $$f(x) = 2x² + 19x + 29$$

The vertex of a quadratic function could be determined by completing the square method to transform the quadratic function from general form to vertex form.

For the general form of a quadratic function, we can do the following transformation.

\begin{aligned} \\f(x)&=ax^2+bx+c\\ & =a(x^2+\dfrac{b}{a}x)+c\\ & =a\Big[x^2+\dfrac{b}{a}x+\Big(\dfrac{b}{2a}\Big)^2-\Big(\dfrac{b}{2a}\Big)^2\Big]+c\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+c-\dfrac{b^2}{4a}\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+\dfrac{4ac-b^2}{4a}\\ \end{aligned}

Therefore, the corrdinates of the vertex is, $$\Big(\dfrac{-b}{2a},\dfrac{4ac-b^2}{4a}\Big)$$

Here we have, $$a=2$$, $$b=19$$ and $$c=29$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-19}{2\cdot 2}\\ & =-\dfrac{19}{4}\\ \end{aligned}

\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 2\cdot29-19^2}{4\cdot2^2}\\ & =-\dfrac{129}{8}\\ \end{aligned}

So the coordinates for the vertex of the quadrautic function are $$\Big(-\dfrac{19}{4},-\dfrac{129}{8}\Big)$$

### 4. Graph for the function $$f(x) = 2x² + 19x + 29$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x² + 19x + 29$$ has two intersection point with the x-axis