(x-4)(x-4)+15(x-4)=0

Solution for (x-4)(x-4)+15(x-4)=0 equation:

(x-4)(x-4)+15(x-4)=0

We multiply parentheses

(x-4)(x-4)+15x-60=0

We multiply parentheses ..

(+x^2-4x-4x+16)+15x-60=0

We get rid of parentheses

x^2-4x-4x+15x+16-60=0

We add all the numbers together, and all the variables

x^2+7x-44=0

a = 1; b = 7; c = -44;
Δ = b2-4ac
Δ = 72-4·1·(-44)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-15}{2*1}=\frac{-22}{2}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+15}{2*1}=\frac{8}{2}
=4
$