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

Solution for 0=(x-5)(5x+2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+5x^2+2x-25x-10))=0


We calculate terms in parentheses: -((+5x^2+2x-25x-10)), so:

(+5x^2+2x-25x-10)

We get rid of parentheses

5x^2+2x-25x-10

We add all the numbers together, and all the variables

5x^2-23x-10

Back to the equation:

-(5x^2-23x-10)


We get rid of parentheses

-5x^2+23x+10=0

a = -5; b = 23; c = +10;
Δ = b2-4ac
Δ = 232-4·(-5)·10
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-27}{2*-5}=\frac{-50}{-10}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+27}{2*-5}=\frac{4}{-10}
=-2/5
$