5(x+4)(x+4)=2(x+1)(x+1)-27

Solution for 5(x+4)(x+4)=2(x+1)(x+1)-27 equation:

5(x+4)(x+4)=2(x+1)(x+1)-27

We move all terms to the left:

5(x+4)(x+4)-(2(x+1)(x+1)-27)=0

We multiply parentheses ..

5(+x^2+4x+4x+16)-(2(x+1)(x+1)-27)=0


We calculate terms in parentheses: -(2(x+1)(x+1)-27), so:

2(x+1)(x+1)-27

We multiply parentheses ..

2(+x^2+x+x+1)-27

We multiply parentheses

2x^2+2x+2x+2-27

We add all the numbers together, and all the variables

2x^2+4x-25

Back to the equation:

-(2x^2+4x-25)


We multiply parentheses

5x^2+20x+20x-(2x^2+4x-25)+80=0

We get rid of parentheses

5x^2-2x^2+20x+20x-4x+25+80=0

We add all the numbers together, and all the variables

3x^2+36x+105=0

a = 3; b = 36; c = +105;
Δ = b2-4ac
Δ = 362-4·3·105
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6}{2*3}=\frac{-42}{6}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6}{2*3}=\frac{-30}{6}
=-5
$