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2y^2+12y-220=0
a = 2; b = 12; c = -220;
Δ = b2-4ac
Δ = 122-4·2·(-220)
Δ = 1904
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1904}=\sqrt{16*119}=\sqrt{16}*\sqrt{119}=4\sqrt{119}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{119}}{2*2}=\frac{-12-4\sqrt{119}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{119}}{2*2}=\frac{-12+4\sqrt{119}}{4} $
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