A=(2x+1)(x-5)+10x-1

Solution for A=(2x+1)(x-5)+10x-1 equation:

=(2A+1)(A-5)+10A-1

We move all terms to the left:

-((2A+1)(A-5)+10A-1)=0

We multiply parentheses ..

-((+2A^2-10A+A-5)+10A-1)=0


We calculate terms in parentheses: -((+2A^2-10A+A-5)+10A-1), so:

(+2A^2-10A+A-5)+10A-1

We get rid of parentheses

2A^2-10A+A+10A-5-1

We add all the numbers together, and all the variables

2A^2+A-6

Back to the equation:

-(2A^2+A-6)


We get rid of parentheses

-2A^2-A+6=0

We add all the numbers together, and all the variables

-2A^2-1A+6=0

a = -2; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-2}=\frac{8}{-4}
=-2
$