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r^2+12r+36=44
We move all terms to the left:
r^2+12r+36-(44)=0
We add all the numbers together, and all the variables
r^2+12r-8=0
a = 1; b = 12; c = -8;
Δ = b2-4ac
Δ = 122-4·1·(-8)
Δ = 176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{11}}{2*1}=\frac{-12-4\sqrt{11}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{11}}{2*1}=\frac{-12+4\sqrt{11}}{2} $
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