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=-18Y^2-19Y+12
We move all terms to the left:
-(-18Y^2-19Y+12)=0
We get rid of parentheses
18Y^2+19Y-12=0
a = 18; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·18·(-12)
Δ = 1225
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{1225}=35$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-35}{2*18}=\frac{-54}{36} =-1+1/2 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+35}{2*18}=\frac{16}{36} =4/9 $
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