121=(2x+3)(2x+3)

Solution for 121=(2x+3)(2x+3) equation:

121=(2x+3)(2x+3)

We move all terms to the left:

121-((2x+3)(2x+3))=0

We multiply parentheses ..

-((+4x^2+6x+6x+9))+121=0


We calculate terms in parentheses: -((+4x^2+6x+6x+9)), so:

(+4x^2+6x+6x+9)

We get rid of parentheses

4x^2+6x+6x+9

We add all the numbers together, and all the variables

4x^2+12x+9

Back to the equation:

-(4x^2+12x+9)


We get rid of parentheses

-4x^2-12x-9+121=0

We add all the numbers together, and all the variables

-4x^2-12x+112=0

a = -4; b = -12; c = +112;
Δ = b2-4ac
Δ = -122-4·(-4)·112
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-44}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+44}{2*-4}=\frac{56}{-8}
=-7
$