(X-5)+x(x+15)=440

Solution for (X-5)+x(x+15)=440 equation:

(X-5)+X(X+15)=440

We move all terms to the left:

(X-5)+X(X+15)-(440)=0

We multiply parentheses

X^2+(X-5)+15X-440=0

We get rid of parentheses

X^2+X+15X-5-440=0

We add all the numbers together, and all the variables

X^2+16X-445=0

a = 1; b = 16; c = -445;
Δ = b2-4ac
Δ = 162-4·1·(-445)
Δ = 2036
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:

$\sqrt{\Delta}=\sqrt{2036}=\sqrt{4*509}=\sqrt{4}*\sqrt{509}=2\sqrt{509}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{509}}{2*1}=\frac{-16-2\sqrt{509}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{509}}{2*1}=\frac{-16+2\sqrt{509}}{2}
$