5x(x-4)+48=2x(x+3)

Solution for 5x(x-4)+48=2x(x+3) equation:

5x(x-4)+48=2x(x+3)

We move all terms to the left:

5x(x-4)+48-(2x(x+3))=0

We multiply parentheses

5x^2-20x-(2x(x+3))+48=0


We calculate terms in parentheses: -(2x(x+3)), so:

2x(x+3)

We multiply parentheses

2x^2+6x

Back to the equation:

-(2x^2+6x)


We get rid of parentheses

5x^2-2x^2-20x-6x+48=0

We add all the numbers together, and all the variables

3x^2-26x+48=0

a = 3; b = -26; c = +48;
Δ = b2-4ac
Δ = -262-4·3·48
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{-(-26)-10}{2*3}=\frac{16}{6}
=2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+10}{2*3}=\frac{36}{6}
=6
$