Steve Nash’s expected value from his one-and-one free throw situation is 1.72 points. What is his free-throw percentage? The one-on-one free throw situation works like this - for the first throw, if you make it, you get to do it again. If you miss, you don't get another chance. If you make it the second time, you get two points total. If Steve Nash's expected value is 1.72 points on average, what is his free throw percentage (number of times he makes the basket)?
Please do not research this on the internet as Steve Nash just retired with the highest free throw percentage, so the answer would be on the internet.
 A: Take x as your free throw probability.
To find the "x" or free throw probability, you need to find out what how many times x goes into 1.72.
To do that, first find what 1.72 is equal to.
The percentage times he would make 2 points would be: 2x^2
Two points would be his free throw percentage (in decimal form) squared because he has to make his free throw twice, (if his free throw percentage as 10% it would be 5% chance of making it twice) * 
The percentage times he would make 1 points would be: x(1-x)
You need to take away x from 1 because that would be the chance that he would miss his second free throw, and multiplied by x for the chance that he will make the first.
The free throw points for 0 would just be left out of the equation.
As 1.72 is equal to the percent of 2 points and the percent of 1 point, 
1.72 = $$
2x^2 + x(1-x)\
$$
$$
1.72 = 2x^2 + x - x^2
$$
$$
1.72 = x^2 + x
$$
$$
0 = x^2 + x - 1.72
$$
We would use the quadratic equation to solve this: 
$$
\frac{-b± sqrt(b^2-4ac)}{2a} \
$$
a = 1
b = 1
c = -1.72
$$
\frac{-1± sqrt(1 + 6.88)}{2} \
$$
$$
\frac{-1± sqrt(7.88)}{2} \
$$
So you solve it:
x = 0.90357
x = 1.90357
Multiply both by 100 to get the percent.
As there is no "190.357%" for this question, the answer has to be 90.357%.
