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2y^2+2y+1=13
We move all terms to the left:
2y^2+2y+1-(13)=0
We add all the numbers together, and all the variables
2y^2+2y-12=0
a = 2; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·2·(-12)
Δ = 100
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{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*2}=\frac{-12}{4} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*2}=\frac{8}{4} =2 $
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