Solve the quadraic equation:

$$x^2-17x-29=0 $$

Quick Answer

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

$$ \Delta=405$$

$$\begin{cases} x_1=\dfrac{17}{2}+\dfrac{9}{2}\sqrt{5} \\ x_2=\dfrac{17}{2}-\dfrac{9}{2}\sqrt{5} \end{cases}$$

In decimal notation,

$$\begin{cases} x_1=18.562305898749 \\ x_2=-1.5623058987491 \end{cases}$$

Detailed Steps on Solution

Solve the quadratic equation: $$x² - 17x - 29 = 0$$

Given $$a =1, b=-17, c=-29$$,

1. Use the quadratic root

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

$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-17)\pm\sqrt{(-17)^2-4\cdot 1\cdot (-29)}}{2 \cdot 1}\\ & =\dfrac{17\pm9\sqrt{5}}{2}\\ & =\dfrac{17}{2}\pm\dfrac{9}{2}\sqrt{5}\\ \end{aligned}$$

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

That is,

$$\begin{cases} x_1 =\dfrac{17}{2}+\dfrac{9}{2}\sqrt{5} \\ x_2=\dfrac{17}{2}-\dfrac{9}{2}\sqrt{5} \end{cases}$$

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,

$$x² - 17x - 29 = 0$$

Move the constant term $$-29$$ to the right hand side. Then its sign becomes postive.


Add square of the half of $$-17$$, the coefficient of the linear term to both sides.


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


Taking square roots on both sides of above equation gives


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


Move the constant $$-\dfrac{17}{2}$$ to the right hand side. Then we get,

$$x_1 = 13$$

$$x_2 = 4$$

3. The vertex of the function $$f(x) = x² - 17x - 29$$

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, $$a=1$$, $$b=-17$$ and $$c=-29$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.

$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-(-17)}{2\cdot 1}\\ & =\dfrac{17}{2}\\ \end{aligned}$$

$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 1\cdot(-29)-(-17)^2}{4\cdot1^2}\\ & =-\dfrac{405}{4}\\ \end{aligned}$$

So the coordinates for the vertex of the quadrautic function are $$\Big(\dfrac{17}{2},-\dfrac{405}{4}\Big)$$

4. Graph for the function $$f(x) = x² - 17x - 29$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x² - 17x - 29$$ has two intersection point with the x-axis

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