(x+3)2=4x+x(x-7)

Solution for (x+3)2=4x+x(x-7) equation:

(x+3)2=4x+x(x-7)

We move all terms to the left:

(x+3)2-(4x+x(x-7))=0

We multiply parentheses

2x-(4x+x(x-7))+6=0


We calculate terms in parentheses: -(4x+x(x-7)), so:

4x+x(x-7)

We multiply parentheses

x^2+4x-7x

We add all the numbers together, and all the variables

x^2-3x

Back to the equation:

-(x^2-3x)


We get rid of parentheses

-x^2+2x+3x+6=0

We add all the numbers together, and all the variables

-1x^2+5x+6=0

a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2}
=-1
$