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=Y2+4Y-5
We move all terms to the left:
-(Y2+4Y-5)=0
We add all the numbers together, and all the variables
-(+Y^2+4Y-5)=0
We get rid of parentheses
-Y^2-4Y+5=0
We add all the numbers together, and all the variables
-1Y^2-4Y+5=0
a = -1; b = -4; c = +5;
Δ = b2-4ac
Δ = -42-4·(-1)·5
Δ = 36
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{36}=6$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*-1}=\frac{-2}{-2} =1 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*-1}=\frac{10}{-2} =-5 $
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