Question

一元二次方程计算：(20t-4t^2)/5=24/25

Answer 1

(20t-4t^2)/5=24/25

5（20t-4t^2)=24

100t-20t^2=24

25t-5t^2-6=0

5t^2-25t+6=0

t^2-5t+6/5=0

(t-5/2)^2-25/4+6/5=0

(t-5/2)^2=101/20

t-5/2=±√505/10

t=5/2±√505/10