(x-8)(2x+10)=(x+2)(x-1)

Solution for (x-8)(2x+10)=(x+2)(x-1) equation:

(x-8)(2x+10)=(x+2)(x-1)

We move all terms to the left:

(x-8)(2x+10)-((x+2)(x-1))=0

We multiply parentheses ..

(+2x^2+10x-16x-80)-((x+2)(x-1))=0


We calculate terms in parentheses: -((x+2)(x-1)), so:

(x+2)(x-1)

We multiply parentheses ..

(+x^2-1x+2x-2)

We get rid of parentheses

x^2-1x+2x-2

We add all the numbers together, and all the variables

x^2+x-2

Back to the equation:

-(x^2+x-2)


We get rid of parentheses

2x^2-x^2+10x-16x-x-80+2=0

We add all the numbers together, and all the variables

x^2-7x-78=0

a = 1; b = -7; c = -78;
Δ = b2-4ac
Δ = -72-4·1·(-78)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-19}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+19}{2*1}=\frac{26}{2}
=13
$