X^2+9x=400=0

Solution for X^2+9x=400=0 equation:

X^2+9X=400=0

We move all terms to the left:

X^2+9X-(400)=0

a = 1; b = 9; c = -400;
Δ = b2-4ac
Δ = 92-4·1·(-400)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-41}{2*1}=\frac{-50}{2}
=-25
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+41}{2*1}=\frac{32}{2}
=16
$