Question

Consider the polynomial equation below.
x2+ x = 20
A. Rewrite the equation so that the right side equals 0.
B. Factor the left side of the equation.
C. Find the roots of the equation.

Answer 1

Find the roots, because when you pass the 20 to the other side (left), the right side would be 0, but you will not have the value of x. So when you pass the 20 to the other side you would have this:
\[ax^{2}+bx+c=0\]

Answer 2

idk i am not good at algebra

Answer 3

To find de roots of the equation you need the quadratic formula:
In this case a=1, b=1, and c=20.
\[x=\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]
\[x_{1}=\frac{ -b+\sqrt{b^{2}-4ac} }{ 2a }\]
\[x_{2}=\frac{ -b-\sqrt{b^{2}-4ac} }{ 2a }\]
just substitute a, b and c on that those equations

Answer 4

the first one is

Answer 5

step one is to write it as \[x^2+ x -20= 0 \]

Answer 6

then use
i think maybe you can factor this one
try
\[(x-4)(x+5)=0\]

Answer 7

that will work because \(5-4=1\) and \(5\times (-4)=-20\)