(x+3)(x+3)=(2x^2+13x+19)

Solution for (x+3)(x+3)=(2x^2+13x+19) equation:

(x+3)(x+3)=(2x^2+13x+19)

We move all terms to the left:

(x+3)(x+3)-((2x^2+13x+19))=0

We multiply parentheses ..

(+x^2+3x+3x+9)-((2x^2+13x+19))=0


We calculate terms in parentheses: -((2x^2+13x+19)), so:

(2x^2+13x+19)

We get rid of parentheses

2x^2+13x+19

Back to the equation:

-(2x^2+13x+19)


We get rid of parentheses

x^2-2x^2+3x+3x-13x+9-19=0

We add all the numbers together, and all the variables

-1x^2-7x-10=0

a = -1; b = -7; c = -10;
Δ = b2-4ac
Δ = -72-4·(-1)·(-10)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*-1}=\frac{4}{-2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*-1}=\frac{10}{-2}
=-5
$