Ah okay so in differential equations you usually want the top variable isolated. To do this, multiply by dt and 2u and you get

Now just integrate both sides. The integral of 2u with respect to u is u². The integral of (2t + sec²(t) with respect to t is t² + ∫sec²(t)dt. The last part is just tan(x) because d/dt(tan(t)) is sec²(t) so just integrating gets us back. Now we have

Where c and k are arbitrary constants. Subtracting c from k and you get

Where b is another constant. To find b, just plug in u(0) = -1 where u is -1 and t is 0. This becomes

tan(0) is 0 so b = 1. Take the plus or minus square root on both sides and you finally get

But Brainly didn't let me do but juat remember there is a plus or minus square root on the left.
Answer:
Attach the image of the triangle so I can help out and edit my answer
Substitute the given into the function
f(-4) = |7(-4) - 3|
now solve
f(-4) = |-28 - 3|
f(-4) = | -31|
the absolute value of |-31| is 31
so,
f(-4) = 31
Hope this helps :)