Question 1180273
Can graph this just to see what it looks like Note that I am assuming b(x)=0.5x+3.25.  I am ignoring the .x" because the formula as I wrote it is a linear equation.
{{{graph(300,300,-1,5,-3,10,-5x^2+14x,0.5x+3.25,4.47,3.38)}}}
Intersect when the two are equal
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so -5x^2+14x=0.5x+3.25
0=5x^2-13.5x+3.25. I like dealing with positive exponents in the square.
Using the quadratic formula where a=5, b=-13.5 and c=3.25
x=(1/10)(13.5+/- sqrt (117.25)); sqrt term=10.83
x=0.267 feet and 2.433 sec or 0.3 sec and 2.4 sec
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h(.267)=3.38 feet substituting .267 into both equations (1 is sufficient, but 2 is a check)
h(2.433)=4.47 feet. Don't round until the last step. Keep all decimal places in x as you calculate. It's much easier to substitute into b(x), since there is no square, but one should check both, and make sure it makes sense on the graph.

The horizontal lines on the graph are at 3.38 and 4.47 feet. They should intersect the graph where the straight line and the curve meet, and they do.