Answer:
Step-by-step explanation:
16t^2-36t+20=0
a = 16; b = -36; c = +20;
Δ = b2-4ac
Δ = -362-4·16·20
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
t1=−b−Δ√2at2=−b+Δ√2a
Δ−−√=16−−√=4
t1=−b−Δ√2a=−(−36)−42∗16=3232=1
t2=−b+Δ√2a=−(−36)+42∗16=4032=1+1/4
