How do you solve #-20x^2 - 20x + 40 = 0#?

2 Answers
Jul 3, 2015

Answer:

You may start by dividing everything by #20#
It also helps if you reverse all the signs.

Explanation:

#->-x^2-x+2=0#
#->x^2+x-2=0#

This can be factorised:
#->(x-1)*(x+2)=0#
#->x=1orx=-2#

Jul 3, 2015

Answer:

I found:
#x_1-=1#
#x_2=-2#

Explanation:

Again, you can use the Quadratic Formula but we can also try something...interesting!
Divide all by #20#;
#-x^2-x+2=0#
Rearrange:
#-x^2-x=-2#
again...
#x^2+x=2#
add and subtract #1/4#;
#x^2+x+1/4-1/4=2#
rearrange:
#x^2+x+1/4=2+1/4#
#(x+1/2)^2=9/4#
root square both sides:
#x+1/2=+-sqrt(9/4)=+-3/2#
so you get two solutions:
#x_1=-1/2+3/2=1#
#x_2=-1/2-3/2=-2#