(x-3)+x+(x+8)(x-5)=360

Solution for (x-3)+x+(x+8)(x-5)=360 equation:

(x-3)+x+(x+8)(x-5)=360

We move all terms to the left:

(x-3)+x+(x+8)(x-5)-(360)=0

We add all the numbers together, and all the variables

x+(x-3)+(x+8)(x-5)-360=0

We get rid of parentheses

x+x+(x+8)(x-5)-3-360=0

We multiply parentheses ..

(+x^2-5x+8x-40)+x+x-3-360=0

We add all the numbers together, and all the variables

(+x^2-5x+8x-40)+2x-363=0

We get rid of parentheses

x^2-5x+8x+2x-40-363=0

We add all the numbers together, and all the variables

x^2+5x-403=0

a = 1; b = 5; c = -403;
Δ = b2-4ac
Δ = 52-4·1·(-403)
Δ = 1637
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{-(5)-\sqrt{1637}}{2*1}=\frac{-5-\sqrt{1637}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1637}}{2*1}=\frac{-5+\sqrt{1637}}{2}
$