13x^2+9(x+1)-(2x+3)(x+2)=6

Solution for 13x^2+9(x+1)-(2x+3)(x+2)=6 equation:

13x^2+9(x+1)-(2x+3)(x+2)=6

We move all terms to the left:

13x^2+9(x+1)-(2x+3)(x+2)-(6)=0

We multiply parentheses

13x^2+9x-(2x+3)(x+2)+9-6=0

We multiply parentheses ..

13x^2-(+2x^2+4x+3x+6)+9x+9-6=0

We add all the numbers together, and all the variables

13x^2-(+2x^2+4x+3x+6)+9x+3=0

We get rid of parentheses

13x^2-2x^2-4x-3x+9x-6+3=0

We add all the numbers together, and all the variables

11x^2+2x-3=0

a = 11; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·11·(-3)
Δ = 136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{34}}{2*11}=\frac{-2-2\sqrt{34}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{34}}{2*11}=\frac{-2+2\sqrt{34}}{22}
$