(x)=(x-5)(2x+1)

Solution for (x)=(x-5)(2x+1) equation:

(x)=(x-5)(2x+1)

We move all terms to the left:

(x)-((x-5)(2x+1))=0

We multiply parentheses ..

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


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

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

We get rid of parentheses

2x^2+x-10x-5

We add all the numbers together, and all the variables

2x^2-9x-5

Back to the equation:

-(2x^2-9x-5)


We add all the numbers together, and all the variables

x-(2x^2-9x-5)=0

We get rid of parentheses

-2x^2+x+9x+5=0

We add all the numbers together, and all the variables

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

a = -2; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·(-2)·5
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*-2}=\frac{-10-2\sqrt{35}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*-2}=\frac{-10+2\sqrt{35}}{-4}
$