(x+1)(14+x+1)=21x^2

Solution for (x+1)(14+x+1)=21x^2 equation:

(x+1)(14+x+1)=21x^2

We move all terms to the left:

(x+1)(14+x+1)-(21x^2)=0

determiningTheFunctionDomain
-21x^2+(x+1)(14+x+1)=0

We add all the numbers together, and all the variables

-21x^2+(x+1)(x+15)=0

We multiply parentheses ..

-21x^2+(+x^2+15x+x+15)=0

We get rid of parentheses

-21x^2+x^2+15x+x+15=0

We add all the numbers together, and all the variables

-20x^2+16x+15=0

a = -20; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·(-20)·15
Δ = 1456
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:

$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{91}}{2*-20}=\frac{-16-4\sqrt{91}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{91}}{2*-20}=\frac{-16+4\sqrt{91}}{-40}
$