Basic Algebraic Manipulation

Question

How would I solve for $X$ in this instance? I can't figure out how to get the $X$ variables by themselves and the known values on the other side by themselves.

$2(4-X)(4-X)+X = 3$

Answer 1

Expand the brackets to get $2(16-8X+X^2)+X=3\Rightarrow 32-16X+2X^2+X-3=0$

$\Rightarrow 2X^2-15X+29=0$, now we use the quadratic equation to get:

$X=\frac{15\pm\sqrt{15^2-4(2)29}}{4}$, simplify these down for your result.

Answer 2

2*(16-4x-4x+x^2)+x = 3

32-16x+2x^2+x = 3

2x^2-16x+32+x = 3

2x^2-15x+32 = 3

2x^2-15x+32-3 = 0

2x^2-15x+29 = 0

Using Bhaskara's formula:

x = (-15 +- sqrt(15^2 - 4*2*29)) / 2*2

x = (-15 +- sqrt(225 - 232) / 4

x = (-15 +- sqrt(-7) / 4

sqrt(-7) gives a complex number, so:

x = (-15 +- i*sqrt(7)) / 4

Answer 3

Hint $\ $ The equation is $\, 2x^2 + x + 1 = 0\ $ for $\ x = X-4.\ $ Solve for $\,x,\,$ then $\, X = x+4.$