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6r^2+102r+432=0
a = 6; b = 102; c = +432;
Δ = b2-4ac
Δ = 1022-4·6·432
Δ = 36
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{36}=6$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-6}{2*6}=\frac{-108}{12} =-9 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+6}{2*6}=\frac{-96}{12} =-8 $
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