sqrt(x^2+5) = 2x-3 solve for x

sqrt(x^2+3) =2x-3

To solve
for x .

Solution:

We can get
rid of the square root  on the left by squaring both sides of the given equation and
get:

x^2+3 = (2x-3)^2

x^2+3 =
(2x)^2+2*(2x)(-3) +(-3)^2, as (a+b)^2 =
a^2+2ab+b^2.

Simplify each term on the
right.

x^2+ 5 =
4x^2-12x+9

Subtract x^2+5  from both
sides.

0 = 4x^2-12x+9 -
x^2-5

0 = 3x^3 -12x+4.

0 =
3x^2-12x +4.

 This is a quadratic equation of the form
ax^2+bx+c  = 0, whose roots are given by:

x1 =
{-b+sqrt(b^2-4ac)}/2a and x2 = {-b-sqrt(b^2-4ac)}/2a

So in
3x^2 - 12x + 2 , a = 3,  b = -12 and  c = 4.

So x1 = (- -12
+sqrt( (-12)^2- 4*3*4)}/(2*3) = (12+sqrt96)/6

x1 =
(6+2sqrt6)/3.

x2 = (6-2sqrt6)/3.