(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+5x-50

Back to the equation:

-(x^2+5x-50)


We get rid of parentheses

2x^2-x^2-4x-7x-5x+14+50=0

We add all the numbers together, and all the variables

x^2-16x+64=0

a = 1; b = -16; c = +64;
Δ = b2-4ac
Δ = -162-4·1·64
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{16}{2}=8$