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12r^2+108r=0
a = 12; b = 108; c = 0;
Δ = b2-4ac
Δ = 1082-4·12·0
Δ = 11664
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{11664}=108$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(108)-108}{2*12}=\frac{-216}{24} =-9 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(108)+108}{2*12}=\frac{0}{24} =0 $
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