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=Y^2+3Y-12
We move all terms to the left:
-(Y^2+3Y-12)=0
We get rid of parentheses
-Y^2-3Y+12=0
We add all the numbers together, and all the variables
-1Y^2-3Y+12=0
a = -1; b = -3; c = +12;
Δ = b2-4ac
Δ = -32-4·(-1)·12
Δ = 57
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*-1}=\frac{3-\sqrt{57}}{-2} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-1}=\frac{3+\sqrt{57}}{-2} $
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