3x^2+12x+13=(2x+5)(2x+5)

Solution for 3x^2+12x+13=(2x+5)(2x+5) equation:

3x^2+12x+13=(2x+5)(2x+5)

We move all terms to the left:

3x^2+12x+13-((2x+5)(2x+5))=0

We multiply parentheses ..

3x^2-((+4x^2+10x+10x+25))+12x+13=0


We calculate terms in parentheses: -((+4x^2+10x+10x+25)), so:

(+4x^2+10x+10x+25)

We get rid of parentheses

4x^2+10x+10x+25

We add all the numbers together, and all the variables

4x^2+20x+25

Back to the equation:

-(4x^2+20x+25)


We add all the numbers together, and all the variables

3x^2+12x-(4x^2+20x+25)+13=0

We get rid of parentheses

3x^2-4x^2+12x-20x-25+13=0

We add all the numbers together, and all the variables

-1x^2-8x-12=0

a = -1; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·(-1)·(-12)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*-1}=\frac{4}{-2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*-1}=\frac{12}{-2}
=-6
$