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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2-6x-1x+3)-((x+5))=0


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

(x+5)

We get rid of parentheses

x+5

Back to the equation:

-(x+5)


We get rid of parentheses

2x^2-6x-1x-x+3-5=0

We add all the numbers together, and all the variables

2x^2-8x-2=0

a = 2; b = -8; c = -2;
Δ = b2-4ac
Δ = -82-4·2·(-2)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{5}}{2*2}=\frac{8-4\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{5}}{2*2}=\frac{8+4\sqrt{5}}{4}
$