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2y-5+2y-5+y2-62=180
We move all terms to the left:
2y-5+2y-5+y2-62-(180)=0
We add all the numbers together, and all the variables
y^2+4y-252=0
a = 1; b = 4; c = -252;
Δ = b2-4ac
Δ = 42-4·1·(-252)
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-32}{2*1}=\frac{-36}{2} =-18 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+32}{2*1}=\frac{28}{2} =14 $
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