NEWTON’S METHOD
Let f(t)=70e⁻¹˙⁵ᵗ+25e⁻⁰˙⁰⁷⁵ᵗ-5,
f'(t)=-105e⁻¹˙⁵ᵗ-1.875e⁻⁰˙⁰⁷⁵ᵗ.
Then, according to Newton’s Method, t=t-f(t)/f'(t) for successive iterations of t.
t₀=21
t₁=21.451356
t₂=21.459170
t₃=21.459172
So the solution to 4 decimal places is t=21.4592.