Consider the equation x2=92+42𝑥2=92+42x 2 =9 2 +4 2 .What value of x𝑥x, to the nearest tenth, makes the equation true where x>0𝑥>0x>0?

The equation you've provided seems to be a bit unclear. However, if you're referring to the quadratic equation in the form of x^2 - 42x + 92 = 0, here's how you can solve it:

Step 1: Identify the coefficients a, b, and c in the equation. In this case, a = 1, b = -42, and c = 92.

Step 2: Use the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Step 3: Substitute the values of a, b, and c into the formula. This gives you x = [42 ± sqrt((-42)^2 - 4192)] / 2*1.

Step 4: Simplify the equation to find the values of x.

Please note that the quadratic formula can give you two solutions, and you're asked to find the solution where x > 0.

If the equation you've provided is different, please clarify so I can provide the correct solution.