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y2-5y=3
We move all terms to the left:
y2-5y-(3)=0
We add all the numbers together, and all the variables
y^2-5y-3=0
a = 1; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·1·(-3)
Δ = 37
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{-(-5)-\sqrt{37}}{2*1}=\frac{5-\sqrt{37}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{37}}{2*1}=\frac{5+\sqrt{37}}{2} $
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