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=Y^2+12Y=33
We move all terms to the left:
-(Y^2+12Y)=0
We get rid of parentheses
-Y^2-12Y=0
We add all the numbers together, and all the variables
-1Y^2-12Y=0
a = -1; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-1)·0
Δ = 144
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{144}=12$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-1}=\frac{0}{-2} =0 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-1}=\frac{24}{-2} =-12 $
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