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12y^2+12y-72=0
a = 12; b = 12; c = -72;
Δ = b2-4ac
Δ = 122-4·12·(-72)
Δ = 3600
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{3600}=60$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-60}{2*12}=\frac{-72}{24} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+60}{2*12}=\frac{48}{24} =2 $
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