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y^2+12y-108=0
a = 1; b = 12; c = -108;
Δ = b2-4ac
Δ = 122-4·1·(-108)
Δ = 576
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*1}=\frac{-36}{2} =-18 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*1}=\frac{12}{2} =6 $
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