The quadratic formula is

$\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Setting $\displaystyle a=1$, $\displaystyle b=5$, $\displaystyle c=\frac{9}{4}$ yields,

$\displaystyle \\ \frac{-5\pm\sqrt{5^2-4(1)(\frac{9}{4})}}{2(1)}\\ \\=\frac{-5\pm\sqrt{25-9}}{2} \\ \\=-\frac{5}{2}\pm2 \\ \\ = -\frac{1}{2}, -\frac{9}{2}$

Identify the values of $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ in the standard form of the parabola.

$\displaystyle y=ax^2+bx+c$

$\displaystyle a= 1$

$\displaystyle b=-1$

$\displaystyle c=100$

Calculate the discriminant.

$\displaystyle b^2-4ac=1-4(1)(100)=-399$

Since the discriminant is less than zero, the quadratic is irreducible and there are no real roots. However, there are complex roots.  Use the quadratic formula to determine the complex roots.

$\displaystyle x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x= \frac{1\pm \sqrt{-399}}{2(1)}$

$\displaystyle x= \frac{1\pm i\sqrt{399}}{2}$

The parent function of a quadratic is represented as $\displaystyle ax^2+bx+c$. The quadratic formula is $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. In this case $\displaystyle a=1$, $\displaystyle b=5$, and $\displaystyle c=-14$. Replacing these values into the quadratic forumula will give you the solutions to the quadratic. 

$\displaystyle x=\frac{-5\pm \sqrt{5^2-4(1)(-14)}}{2*1}$

$\displaystyle x=\frac{-5\pm \sqrt{25-(-56)}}{2}$

$\displaystyle x=\frac{-5\pm \sqrt{25-(-56)}}{2}$

$\displaystyle x=\frac{-5\pm \sqrt{81}}{2}$

$\displaystyle x=\frac{-5\pm 9}{2}$

$\displaystyle x=-7$ and $\displaystyle x=2$

You must know the quadratic equation $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2}$.

To plug in the right terms, recognize that polynomials in standard form are symbolized as $\displaystyle ax^2+bx+c$.

Plug in the values from your equation

$\displaystyle x=\frac{-4\pm \sqrt{4^2-4*1*2}}{2}$

simplify within the radical:

$\displaystyle x=\frac{-4\pm \sqrt{8}}{2}$

Simplify the radical:

$\displaystyle x=\frac{-4\pm2 \sqrt{2}}{2}$

Reduce:

$\displaystyle x={-2\pm \sqrt{2}$

Note that this represents two values since there is a $\displaystyle \pm$ in the equation. One is solved with an addition sign and the other is solved with a subtraction sign to yield two answers or roots where this equation crosses the x axis. 

1) Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

$\displaystyle x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\displaystyle x=\frac{4\pm \sqrt{(-4)^{2}-4(1)(-4)}}{2\cdot (1)}$

$\displaystyle x=\frac{4\pm \sqrt{16+16}}{2}$

$\displaystyle x=\frac{4\pm \sqrt{32}}{2}$

$\displaystyle x=\frac{4\pm \sqrt{16\cdot 2}}{2}=\frac{4\pm \sqrt{16}\cdot \sqrt{2}}{2}=\frac{4\pm4\sqrt{2}}{2}$

$\displaystyle x=2\pm2\sqrt{2}$

The standard form of a quadratic equation is $\displaystyle ax^{2}+bx+c=0$, where a,b, and c are constants. Plug these constants into the quadratic formula to solve for x.

$\displaystyle x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\displaystyle x=\frac{-12\pm \sqrt{(12)^{2}-4(2)(15)}}{2(2)}$

$\displaystyle x=\frac{-12\pm \sqrt{144-120}}{4}=\frac{-12\pm \sqrt{24}}{4}$

$\displaystyle x=\frac{-12\pm 2\sqrt{6}}{4}$$\displaystyle x=-3\pm \frac{\sqrt{6}}2{}$

The quadratic equations is:

$\displaystyle x=\frac{-b\pm{\sqrt{b^2-4ac}}{}}{2a}$

From here you just plug in your numbers, so:

$\displaystyle x=\frac{-14\pm{\sqrt{14^2-4(1)(2)}}{}}{2(1)}$

Simplify:

$\displaystyle \frac{-14\pm \sqrt{188}}{2}$

Then you need to simplify the inside looking for perfect squares:

$\displaystyle \frac{-14\pm \sqrt{4}*\sqrt{47}}{2}=\frac{-14\pm 2\sqrt{47}}{2}$

Each term is divisible by 2, so your final answer is:

$\displaystyle {-7\pm \sqrt{47}$

$\displaystyle a=3$, $\displaystyle b=-7$, $\displaystyle c=11$

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x=\frac{7\pm \sqrt{(-7)^2-4(3)(11)}}{2(3)}$

$\displaystyle x=\frac{7\pm \sqrt{49-132}}{6}$

$\displaystyle x=\frac{7\pm \sqrt{-83}}{6}$

$\displaystyle x=\frac{7\pm i\sqrt{83}}{6}$

This specific function cannot be factored, so use the quadratic equation:

$\displaystyle \frac{-b\pm \sqrt{b^2 -4ac}}{2a}$

Our function is in the form $\displaystyle ax^2+bx+c$ where, 

$\displaystyle a=2, b=3, c=1$

Therefore the quadratic equation becomes,

 $\displaystyle \frac{-3\pm \sqrt{3^2 -4(2)(1)}}{2(2)}$

$\displaystyle \frac{-3\pm \sqrt{9 -8}}{4}$

$\displaystyle \frac{-3\pm \sqrt{1}}{4}$

$\displaystyle \frac{-3+1}{4}$ OR $\displaystyle \frac{-3-1}{4}$

$\displaystyle \frac{-2}{4}$ OR $\displaystyle \frac{-4}{4}$

$\displaystyle x=-\frac{1}{2}$  OR  $\displaystyle x=-1$

Use the quadratic equation: 

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Since the original equation is in standard form, $\displaystyle ax^2+bx+c$ where 

$\displaystyle a=2, b=4, c=4$.

Therefore,

$\displaystyle x=\frac{-4\pm\sqrt{4^2-4(2)(4)}}{2(2)}$

$\displaystyle x=\frac{-4\pm\sqrt{16-32}}{4}$

$\displaystyle x=\frac{-4\pm\sqrt{-16}}{4}$

$\displaystyle x=\frac{-4\pm4i}{4}$

$\displaystyle x=-1 \pm i$

because the value of the discriminant (the component beneath the square root) is negative, this function has no real solutions.