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x2+9x=400
We move all terms to the left:
x2+9x-(400)=0
We add all the numbers together, and all the variables
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 $
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