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y^2+12y=-5
We move all terms to the left:
y^2+12y-(-5)=0
We add all the numbers together, and all the variables
y^2+12y+5=0
a = 1; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·1·5
Δ = 124
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{31}}{2*1}=\frac{-12-2\sqrt{31}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{31}}{2*1}=\frac{-12+2\sqrt{31}}{2} $
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