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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+x^2-5x+3x-15)-(2x(x+4))=0


We calculate terms in parentheses: -(2x(x+4)), so:

2x(x+4)

We multiply parentheses

2x^2+8x

Back to the equation:

-(2x^2+8x)


We get rid of parentheses

x^2-2x^2-5x+3x-8x-15=0

We add all the numbers together, and all the variables

-1x^2-10x-15=0

a = -1; b = -10; c = -15;
Δ = b2-4ac
Δ = -102-4·(-1)·(-15)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{10}}{2*-1}=\frac{10-2\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{10}}{2*-1}=\frac{10+2\sqrt{10}}{-2}
$