How do I set (2x+5)(2x+5)=300 to zero

Hi,
I am not entirely sure that I understood what you mean but it's probably:
(2X+5)(2X+5)=300
4x^2 + 20X + 25 =300
4X^2 + 20X - 275 = 0
Best regards,

You rewrite it as

(2x+5)(2x+5) - 300 = 4x^2 + 20x - 275 = 0

This allows you to extract the roots of the equation from

x = - 20 +/-  [20 ^2  - 4 x 4 x (- 275) ]^(1/2)

   = - 20 +/- (400 + 4,400)^(1/2) = -20 +/-  4,800^(1/2)

   =  -20 +/-  69.282032.

Therefore the roots of the equation are:

x = 49.282032   and  x = - 89.282032

These roots set the equation to zero because they allow you to rewrite it

as

(x - 49.282032)(x + 89.282032) = 0.

Erratum: In the preceding answer by Harry the roots of the quadratic equation should be divided by 8. Thus

x = 6.160254  and  x = -11.160254

Therefore the quadratic equation is set to zero by rewriting it as

(x - 6.160254)(x + 11.160254) = 0

(2x + 5)(2x +5) = 300
To set it to zero, you just subtract 300 from both sides, resulting in this:
(2x + 5)(2x +5) - 300 = 0
Here is the rest of the solution:
Multiply out the (2x + 5)(2x +5) part using the FOIL method, or however you know how to do it, if you don’t know the method by name:
(4x^2 + 10x + 10x +25) - 300 = 0
Then remove parentheses and combine like terms, in this case the whole numbers 25 and -300, and the 10x and 10x:
4x^2 + 20x - 275 = 0
Then use the quadratic formula to solve for x:

x = -b +/- sq. rt. (b^2- 4ac) 
                   2a

x = -20 +/- sq. rt. [20^2 - (4*4*-275)]   

                         2*4

=   -20 +/- sq. rt. [400 - ( -4400)] 
                         8
 
= -20 +/ sq. rt. (4800)    =    -20 +/- 69.28204    
                8                                    8

So your 2 answers are x = -20 + 69.28204    =   6.160255     and
                                              8 
                                x = -20 - 69.28204     = - 11.160255
                                               8