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y2-3y=6
We move all terms to the left:
y2-3y-(6)=0
We add all the numbers together, and all the variables
y^2-3y-6=0
a = 1; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·1·(-6)
Δ = 33
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{33}}{2*1}=\frac{3-\sqrt{33}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*1}=\frac{3+\sqrt{33}}{2} $
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