-3x-5=(x+3)(x+3)

Solution for -3x-5=(x+3)(x+3) equation:

-3x-5=(x+3)(x+3)

We move all terms to the left:

-3x-5-((x+3)(x+3))=0

We multiply parentheses ..

-((+x^2+3x+3x+9))-3x-5=0


We calculate terms in parentheses: -((+x^2+3x+3x+9)), so:

(+x^2+3x+3x+9)

We get rid of parentheses

x^2+3x+3x+9

We add all the numbers together, and all the variables

x^2+6x+9

Back to the equation:

-(x^2+6x+9)


We add all the numbers together, and all the variables

-3x-(x^2+6x+9)-5=0

We get rid of parentheses

-x^2-3x-6x-9-5=0

We add all the numbers together, and all the variables

-1x^2-9x-14=0

a = -1; b = -9; c = -14;
Δ = b2-4ac
Δ = -92-4·(-1)·(-14)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*-1}=\frac{4}{-2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*-1}=\frac{14}{-2}
=-7
$