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16b^2+96b=0
a = 16; b = 96; c = 0;
Δ = b2-4ac
Δ = 962-4·16·0
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9216}=96$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-96}{2*16}=\frac{-192}{32} =-6 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+96}{2*16}=\frac{0}{32} =0 $
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