9X(x-1)+(x-4)=16

Solution for 9X(x-1)+(x-4)=16 equation:

9X(X-1)+(X-4)=16

We move all terms to the left:

9X(X-1)+(X-4)-(16)=0

We multiply parentheses

9X^2-9X+(X-4)-16=0

We get rid of parentheses

9X^2-9X+X-4-16=0

We add all the numbers together, and all the variables

9X^2-8X-20=0

a = 9; b = -8; c = -20;
Δ = b2-4ac
Δ = -82-4·9·(-20)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-28}{2*9}=\frac{-20}{18}
=-1+1/9
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+28}{2*9}=\frac{36}{18}
=2
$