(x-3)(2x+1)=x(x+5)

Solution for (x-3)(2x+1)=x(x+5) equation:

(x-3)(2x+1)=x(x+5)

We move all terms to the left:

(x-3)(2x+1)-(x(x+5))=0

We multiply parentheses ..

(+2x^2+x-6x-3)-(x(x+5))=0


We calculate terms in parentheses: -(x(x+5)), so:

x(x+5)

We multiply parentheses

x^2+5x

Back to the equation:

-(x^2+5x)


We get rid of parentheses

2x^2-x^2+x-6x-5x-3=0

We add all the numbers together, and all the variables

x^2-10x-3=0

a = 1; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·1·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{7}}{2*1}=\frac{10-4\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{7}}{2*1}=\frac{10+4\sqrt{7}}{2}
$