(5x-6)(x-5)=2(x+5)+5(x-4)

Solution for (5x-6)(x-5)=2(x+5)+5(x-4) equation:

(5x-6)(x-5)=2(x+5)+5(x-4)

We move all terms to the left:

(5x-6)(x-5)-(2(x+5)+5(x-4))=0

We multiply parentheses ..

(+5x^2-25x-6x+30)-(2(x+5)+5(x-4))=0


We calculate terms in parentheses: -(2(x+5)+5(x-4)), so:

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

We multiply parentheses

2x+5x+10-20

We add all the numbers together, and all the variables

7x-10

Back to the equation:

-(7x-10)


We get rid of parentheses

5x^2-25x-6x-7x+30+10=0

We add all the numbers together, and all the variables

5x^2-38x+40=0

a = 5; b = -38; c = +40;
Δ = b2-4ac
Δ = -382-4·5·40
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{161}}{2*5}=\frac{38-2\sqrt{161}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{161}}{2*5}=\frac{38+2\sqrt{161}}{10}
$