(2x+3)(2x+3)+(2x+3)(7x-2)=0

Solution for (2x+3)(2x+3)+(2x+3)(7x-2)=0 equation:

(2x+3)(2x+3)+(2x+3)(7x-2)=0

We multiply parentheses ..

(+4x^2+6x+6x+9)+(2x+3)(7x-2)=0

We get rid of parentheses

4x^2+6x+6x+(2x+3)(7x-2)+9=0

We multiply parentheses ..

4x^2+(+14x^2-4x+21x-6)+6x+6x+9=0

We add all the numbers together, and all the variables

4x^2+(+14x^2-4x+21x-6)+12x+9=0

We get rid of parentheses

4x^2+14x^2-4x+21x+12x-6+9=0

We add all the numbers together, and all the variables

18x^2+29x+3=0

a = 18; b = 29; c = +3;
Δ = b2-4ac
Δ = 292-4·18·3
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-25}{2*18}=\frac{-54}{36}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+25}{2*18}=\frac{-4}{36}
=-1/9
$