(x-3)(x-3)+15(x-4)=0

Solution for (x-3)(x-3)+15(x-4)=0 equation:

(x-3)(x-3)+15(x-4)=0

We multiply parentheses

(x-3)(x-3)+15x-60=0

We multiply parentheses ..

(+x^2-3x-3x+9)+15x-60=0

We get rid of parentheses

x^2-3x-3x+15x+9-60=0

We add all the numbers together, and all the variables

x^2+9x-51=0

a = 1; b = 9; c = -51;
Δ = b2-4ac
Δ = 92-4·1·(-51)
Δ = 285
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{285}}{2*1}=\frac{-9-\sqrt{285}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{285}}{2*1}=\frac{-9+\sqrt{285}}{2}
$