(3x+2)(3x+2)+19=100

Solution for (3x+2)(3x+2)+19=100 equation:

(3x+2)(3x+2)+19=100

We move all terms to the left:

(3x+2)(3x+2)+19-(100)=0

We add all the numbers together, and all the variables

(3x+2)(3x+2)-81=0

We multiply parentheses ..

(+9x^2+6x+6x+4)-81=0

We get rid of parentheses

9x^2+6x+6x+4-81=0

We add all the numbers together, and all the variables

9x^2+12x-77=0

a = 9; b = 12; c = -77;
Δ = b2-4ac
Δ = 122-4·9·(-77)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-54}{2*9}=\frac{-66}{18}
=-3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+54}{2*9}=\frac{42}{18}
=2+1/3
$