(x+3)(x-3)+(2x+3)*2=5x*2+7(x-5)

Solution for (x+3)(x-3)+(2x+3)*2=5x*2+7(x-5) equation:

(x+3)(x-3)+(2x+3)*2=5x*2+7(x-5)

We move all terms to the left:

(x+3)(x-3)+(2x+3)*2-(5x*2+7(x-5))=0

We use the square of the difference formula

x^2+(2x+3)*2-(5x*2+7(x-5))-9=0

We multiply parentheses

x^2+4x-(5x*2+7(x-5))+6-9=0


We calculate terms in parentheses: -(5x*2+7(x-5)), so:

5x*2+7(x-5)

We multiply parentheses

5x*2+7x-35

Wy multiply elements

10x+7x-35

We add all the numbers together, and all the variables

17x-35

Back to the equation:

-(17x-35)


We add all the numbers together, and all the variables

x^2+4x-(17x-35)-3=0

We get rid of parentheses

x^2+4x-17x+35-3=0

We add all the numbers together, and all the variables

x^2-13x+32=0

a = 1; b = -13; c = +32;
Δ = b2-4ac
Δ = -132-4·1·32
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{41}}{2*1}=\frac{13-\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{41}}{2*1}=\frac{13+\sqrt{41}}{2}
$