# ((2X-14Cos45)/(49+X^2-14XCos45))+0.3536 = 0

Solve to find the value of X

Change cos 45 to sqrt 2 and rewrite the equation in the form
a x^2 + bx + c = 0

Start out by rewriting it as
2x - 9.8995
= (-0.3536)(49 + x^2 - 9.8995 x)

Conimue until you get the form
a x^2 + bx + c = 0, figure out what a, b, and c are, and use the quadratic equation for the two different values of x that are the roots of the equation.

If you are unfamiliar with the quadratic equation method, review

Sorry, I made a mistake in my previous answer. Cos 45 should have been (1/2) sqrt 2 = 0.70711.., not sqrt 2. Duh. The procedure is OK; the numbers are wrong.

## To solve the equation ((2X-14Cos45)/(49+X^2-14XCos45))+0.3536 = 0, we need to first simplify the equation and then solve for X.

First, we can replace cos 45 with its value, which is (1/2) √2 or approximately 0.70711. The equation becomes:

((2X - 14 * 0.70711) / (49 + X^2 - 14X * 0.70711)) + 0.3536 = 0.

Next, let's simplify the numerator and denominator separately:

Numerator: 2X - 14 * 0.70711 = 2X - 9.8995.

Denominator: 49 + X^2 - 14X * 0.70711 = 49 + X^2 - 9.8995X.

Now, let's substitute these simplified expressions back into the equation:

(2X - 9.8995) / (49 + X^2 - 9.8995X) + 0.3536 = 0.

To further simplify, let's multiply both sides of the equation by the denominator:

(2X - 9.8995) + 0.3536(49 + X^2 - 9.8995X) = 0.

Expanding and rearranging the terms:

2X - 9.8995 + 0.3536(49) + 0.3536(X^2) - 0.3536(9.8995X) = 0.

2X - 9.8995 + 17.3524 + 0.3536X^2 - 3.4996X = 0.

Collecting like terms:

0.3536X^2 - 1.4996X + 7.4529 = 0.

Now, we have the equation in the form aX^2 + bX + c = 0, where a = 0.3536, b = -1.4996, and c = 7.4529.

X = (-b ± √(b^2 - 4ac)) / (2a).

Plugging in the values of a, b, and c:

X = (-(-1.4996) ± √((-1.4996)^2 - 4 * 0.3536 * 7.4529)) / (2 * 0.3536).

Simplifying further:

X = (1.4996 ± √(2.2492 - 10.515712)) / 0.7072.

X = (1.4996 ± √(-8.266512)) / 0.7072.

Since the discriminant (√(b^2 - 4ac)) is negative, there are no real solutions to the equation.