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x2+2x-99=0
We add all the numbers together, and all the variables
x^2+2x-99=0
a = 1; b = 2; c = -99;
Δ = b2-4ac
Δ = 22-4·1·(-99)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-20}{2*1}=\frac{-22}{2} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+20}{2*1}=\frac{18}{2} =9 $
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