(2x-7)(x-2)=(x-10)(x-5)

Solution for (2x-7)(x-2)=(x-10)(x-5) equation:

(2x-7)(x-2)=(x-10)(x-5)

We move all terms to the left:

(2x-7)(x-2)-((x-10)(x-5))=0

We multiply parentheses ..

(+2x^2-4x-7x+14)-((x-10)(x-5))=0


We calculate terms in parentheses: -((x-10)(x-5)), so:

(x-10)(x-5)

We multiply parentheses ..

(+x^2-5x-10x+50)

We get rid of parentheses

x^2-5x-10x+50

We add all the numbers together, and all the variables

x^2-15x+50

Back to the equation:

-(x^2-15x+50)


We get rid of parentheses

2x^2-x^2-4x-7x+15x+14-50=0

We add all the numbers together, and all the variables

x^2+4x-36=0

a = 1; b = 4; c = -36;
Δ = b2-4ac
Δ = 42-4·1·(-36)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*1}=\frac{-4-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*1}=\frac{-4+4\sqrt{10}}{2}
$