x(x-4)=3x(x+6)+48

Solution for x(x-4)=3x(x+6)+48 equation:

x(x-4)=3x(x+6)+48

We move all terms to the left:

x(x-4)-(3x(x+6)+48)=0

We multiply parentheses

x^2-4x-(3x(x+6)+48)=0


We calculate terms in parentheses: -(3x(x+6)+48), so:

3x(x+6)+48

We multiply parentheses

3x^2+18x+48

Back to the equation:

-(3x^2+18x+48)


We get rid of parentheses

x^2-3x^2-4x-18x-48=0

We add all the numbers together, and all the variables

-2x^2-22x-48=0

a = -2; b = -22; c = -48;
Δ = b2-4ac
Δ = -222-4·(-2)·(-48)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*-2}=\frac{12}{-4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*-2}=\frac{32}{-4}
=-8
$