(3x+5)(3x+5)=(x+1)(x+1)

Solution for (3x+5)(3x+5)=(x+1)(x+1) equation:

(3x+5)(3x+5)=(x+1)(x+1)

We move all terms to the left:

(3x+5)(3x+5)-((x+1)(x+1))=0

We multiply parentheses ..

(+9x^2+15x+15x+25)-((x+1)(x+1))=0


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

(x+1)(x+1)

We multiply parentheses ..

(+x^2+x+x+1)

We get rid of parentheses

x^2+x+x+1

We add all the numbers together, and all the variables

x^2+2x+1

Back to the equation:

-(x^2+2x+1)


We get rid of parentheses

9x^2-x^2+15x+15x-2x+25-1=0

We add all the numbers together, and all the variables

8x^2+28x+24=0

a = 8; b = 28; c = +24;
Δ = b2-4ac
Δ = 282-4·8·24
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{-(28)-4}{2*8}=\frac{-32}{16}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*8}=\frac{-24}{16}
=-1+1/2
$