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y2-12y-36=0
We add all the numbers together, and all the variables
y^2-12y-36=0
a = 1; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·1·(-36)
Δ = 288
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{2}}{2*1}=\frac{12-12\sqrt{2}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{2}}{2*1}=\frac{12+12\sqrt{2}}{2} $
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