SOLUTION: solve for x: 80x^2-96x+27=0
Algebra.Com
Question 1069121: solve for x: 80x^2-96x+27=0
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
80x^2-96x+27=0
x=(1/160)(96+/- sqrt (96^2-8640); sqrt (9216-8640)=sqrt (576)=24
x=(1/160)(96+/-24)
x=(1/160)(120), or 3/4
x=(1/160)(72) or 9/20
80(9/16)-96(3/4)+27=45-72+27=0
80(.2025)-96(.45)+27=16.2-43.2+27=0
RELATED QUESTIONS
Solve for X:... (answered by jim_thompson5910)
solve the inequality using interval notation for x^4 + 5x^3 -16x^2-80x... (answered by lwsshak3)
solve by factoring.... (answered by addingup)
Solve the equation.... (answered by Alan3354)
-16x^2+80x+5=0... (answered by Edwin McCravy)
x^4-20x^3+96x^2 could you help me solve this problem on factoring... (answered by jim_thompson5910)
x^2 + 6x – 27 = 0 solve for... (answered by rwm)
I need to use Rational Zero Theorum to solve x^4-5x^3-48x^2-80x-24=0
Also what is the... (answered by Alan3354)
Solve for x:... (answered by mananth)