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2y^2+22y+60=0
a = 2; b = 22; c = +60;
Δ = b2-4ac
Δ = 222-4·2·60
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*2}=\frac{-24}{4} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*2}=\frac{-20}{4} =-5 $
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