(x+1)(x)=11(2x+1)+35

Solution for (x+1)(x)=11(2x+1)+35 equation:

(x+1)(x)=11(2x+1)+35

We move all terms to the left:

(x+1)(x)-(11(2x+1)+35)=0

We multiply parentheses

x^2+x-(11(2x+1)+35)=0


We calculate terms in parentheses: -(11(2x+1)+35), so:

11(2x+1)+35

We multiply parentheses

22x+11+35

We add all the numbers together, and all the variables

22x+46

Back to the equation:

-(22x+46)


We get rid of parentheses

x^2+x-22x-46=0

We add all the numbers together, and all the variables

x^2-21x-46=0

a = 1; b = -21; c = -46;
Δ = b2-4ac
Δ = -212-4·1·(-46)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-25}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+25}{2*1}=\frac{46}{2}
=23
$