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y2-3y-99=0
We add all the numbers together, and all the variables
y^2-3y-99=0
a = 1; b = -3; c = -99;
Δ = b2-4ac
Δ = -32-4·1·(-99)
Δ = 405
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9\sqrt{5}}{2*1}=\frac{3-9\sqrt{5}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9\sqrt{5}}{2*1}=\frac{3+9\sqrt{5}}{2} $
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