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

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

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

We move all terms to the left:

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

We multiply parentheses

x^2+4x+3x-(7(x+4))+12=0


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

7(x+4)

We multiply parentheses

7x+28

Back to the equation:

-(7x+28)


We add all the numbers together, and all the variables

x^2+7x-(7x+28)+12=0

We get rid of parentheses

x^2+7x-7x-28+12=0

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$