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y^2+12y-120=0
a = 1; b = 12; c = -120;
Δ = b2-4ac
Δ = 122-4·1·(-120)
Δ = 624
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{39}}{2*1}=\frac{-12-4\sqrt{39}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{39}}{2*1}=\frac{-12+4\sqrt{39}}{2} $
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