Question

Use quadratic formula to solve the equation
-x^2= 16x+46

Answer 1

\(\bf -x^2= 16x+46\implies -1x^2-16x-46=0\implies 1x^2+16x+46=0 \\ \quad \\
\textit{quadratic formula}\\
\begin{array}{llll}
y=&ax^2+&bx+&c\\
&\uparrow&\ \uparrow&\uparrow\\
&a&\ b&c\\
&1x^2&+16x&+46
\end{array}
x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\)

Answer 2

How did you get positive 16 and 46

Answer 3

well, you can use the negative version, just the same :), it''d be just a bit cleaner using the positive one, but \(\bf -x^2= 16x+46\implies -1x^2-16x-46=0 \quad \times -1\textit{ both sides}\\ \quad \\
\implies 1x^2+16x+46=0\)

Answer 4

recall that -1 * 0 = 0

Answer 5

\[-x^2=16x+46\]Rearrange to the standard \[ax^2+bx+c=0\]by adding \(x^2\) to each side:\[-x^2+x^2=x^2+16x+46\]\[0=x^2+16x+46\]\[a=1,\,b=16,\,c=46\]and you're off to the races