Using the quadratic formula, with $\displaystyle a = 1, b = 4, c = -17$:

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

$\displaystyle x = \frac{-(4) \pm \sqrt{(4)^{2}-4 (1)(-17)}}{2 (1)}$

$\displaystyle x = \frac{-4 \pm \sqrt{16-(-68)}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{84}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{4} \cdot \sqrt{21}}{2 }$

$\displaystyle x = \frac{-4 \pm 2 \sqrt{21}}{2 }$

$\displaystyle x = -2 \pm \sqrt{21}$

Using the quadratic formula, with $\displaystyle a = 1, b = 4, c = 13$:

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

$\displaystyle x = \frac{-(4) \pm \sqrt{(4)^{2}-4 (1)(13)}}{2 (1)}$

$\displaystyle x = \frac{-4 \pm \sqrt{16-52}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{-36}}{2 }$

$\displaystyle x = \frac{-4 \pm i \sqrt{36}}{2 }$

$\displaystyle x = \frac{-4 \pm 6 i }{2 }$

$\displaystyle x = -2 \pm 3 i$

Write in the form $\displaystyle (x-p)(x-q)=0$ where p and q are the roots.

Substitute in the roots:

$\displaystyle (x-2)(x-(-10))=0$

Simplify:

$\displaystyle (x-2)(x+10)=0$

Use FOIL and simplify to get

$\displaystyle x^2+8x-20=0$.

To find the roots of this equation, we need to find which values of $\displaystyle x$ make the polynomial equal zero; we do this by factoring. Factoring is a lot of "guess and check" work, but we can figure some things out. If our binomials are in the form $\displaystyle (ax+b)(cx+d)$, we know $\displaystyle a$ times $\displaystyle c$ will be $\displaystyle 6$ and $\displaystyle b$ times $\displaystyle d$ will be $\displaystyle -28$. With that in mind, we can factor our polynomial to 

$\displaystyle (2x-4)(3x+7)$

To find the roots, we need to find the $\displaystyle x$-values that make each of our binomials equal zero. For the first one it is $\displaystyle 2$, and for the second it is $\displaystyle -\frac{7}{3}$, so our roots are $\displaystyle 2, -\frac{7}{3}$.

1. Write the equation in the form $\displaystyle (x-a)(x-b)=0$ where $\displaystyle a$ and $\displaystyle b$ are the given roots.

$\displaystyle (x-4)(x-(-8))=0$

$\displaystyle (x-4)(x+8)=0$

2. Simplify using FOIL method.

$\displaystyle x^{2}+4x-32=0$

Use the quadratic formula with $\displaystyle a=1$, $\displaystyle b=3$ and $\displaystyle c=3$:

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

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

$\displaystyle x=\frac{-3\pm \sqrt{9-12}}{2}$

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

$\displaystyle x=\frac{-3\pm i\sqrt{3}}{2}$

Use the quadratic formula with $\displaystyle a=10$, $\displaystyle b=4$, and $\displaystyle c=5$:

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

$\displaystyle x=\frac{-4\pm \sqrt{4^{2}-4(10)(5)}}{2(10)}$

$\displaystyle x=\frac{-4\pm \sqrt{16-200}}{20}$

$\displaystyle x=\frac{-4\pm \sqrt{-184}}{20}$

$\displaystyle x=\frac{-4\pm \sqrt{46\cdot 4\cdot -1}}{20}$

$\displaystyle x=\frac{-4\pm 2i\sqrt{46}}{20}$

$\displaystyle x=\frac{-2\pm i\sqrt{46}}{10}$

To solve for x we need to isolate x. We can do this by taking the square root of each side and then doing algebraic operations.

$\displaystyle (x-3)^2=36$

$\displaystyle \sqrt(x-3)^2=\sqrt36$

$\displaystyle x-3=\pm6$

Now we need to separate our equation in two and solve for each x.

$\displaystyle x-3=6$     or     $\displaystyle x-3=-6$

$\displaystyle x=9$                     $\displaystyle x=-3$

To solve this equation, you must first eliminate the exponent from the $\displaystyle (x-3)$ by taking the square root of both sides: 

$\displaystyle \sqrt{(x-3)^{2}}=\sqrt{36}$

Since the square root of 36 could be either $\displaystyle 6$ or $\displaystyle -6$, there must be 2 values of $\displaystyle x$. So, solve for

$\displaystyle x-3=-6$

and

$\displaystyle x-3=6$

to get solutions of $\displaystyle (9,-3)$.

When we factor, we are looking for two number that multiply to the constant, $\displaystyle 14$, and add to the middle term, $\displaystyle 9$. Looking through the factors of $\displaystyle 14$, we can find those factors to be $\displaystyle 7$ and $\displaystyle 2$.

Thus, we have the factors: 

$\displaystyle (x + 2)(x+7)$.

To solve for the solutions, set each of these factors equal to zero.

Thus, we get $\displaystyle x + 2 = 0$, or $\displaystyle x = -2$.

Our second solution is, $\displaystyle x + 7 =0$, or $\displaystyle x = -7$.