x2+40x+300=225

Solution for x2+40x+300=225 equation:

x2+40x+300=225

We move all terms to the left:

x2+40x+300-(225)=0

We add all the numbers together, and all the variables

x^2+40x+75=0

a = 1; b = 40; c = +75;
Δ = b2-4ac
Δ = 402-4·1·75
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{13}}{2*1}=\frac{-40-10\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{13}}{2*1}=\frac{-40+10\sqrt{13}}{2}
$