solve 300=-5x^2+100x+20 by factoring

[LaryssaLacerda]

I am stuck on the first part.. I can't seem to figure out if I have d\to do factoring

300 = -5x^2 + 100x + 20

300 - 20 = -5x^2 + 100x + 20 - 20

280 = -5x^2 + 100x

280 - 280 = -5x^2 + 100x - 280

0 = -5x^2 + 100x - 280

0 = (-5x^2 + 100x - 280) / (-5)

0 = x^2 - 20x + 56

 

[soroban]

Re: again

Hello, Laryssa!

\(\displaystyle 300\:=\:-5x^2\,+\,100x\,+\,20\)

\(\displaystyle 300\,-\,20\:=\:-5x^2\,+\,100x\,+\,20\,-\,20\)

\(\displaystyle 280\:=\:-5x^2\,+\,100x\)

\(\displaystyle 280\,-\,280\:=\:-5x^2\,+\,100x\,-\,280\)

\(\displaystyle 0\:=\:-5x^2\,+\,100x\,-\,280\)

\(\displaystyle 0\:=\:\frac{-5x^2\,+\,100x\,-\,280}{-5}\)

\(\displaystyle 0\:=\:x^2\,-\,20x\,+\,56\) . . . What was all that?

We have: \(\displaystyle \:300 \:=\:-5x^2\,+\,100x\,+\,20\)

Bring all terms to the left: \(\displaystyle \:5x^2\,-\,100x\,+\,280\:=\:0\)

Divide by 5: \(\displaystyle \:x^2\,-\,20x \,+\,56\:=\:0\)


No, it doesn't factor . . . We must use the Quadratic Formula!

\(\displaystyle x\:=\:\frac{20\,\pm\,\sqrt{20^2\,-\,4(1)(56)}}{2(1)}\:=\:\frac{20\,\pm\,\sqrt{176}}{2} \:=\:\frac{20\,\pm\,4\sqrt{11}}{2}\)

Therefore: \(\displaystyle \L\:x\:=\:10\,\pm\,2\sqrt{11}\)