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y2+3y=1
We move all terms to the left:
y2+3y-(1)=0
We add all the numbers together, and all the variables
y^2+3y-1=0
a = 1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·1·(-1)
Δ = 13
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{13}}{2*1}=\frac{-3-\sqrt{13}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*1}=\frac{-3+\sqrt{13}}{2} $
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