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4p^2+44p^2+96p=0
We add all the numbers together, and all the variables
48p^2+96p=0
a = 48; b = 96; c = 0;
Δ = b2-4ac
Δ = 962-4·48·0
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9216}=96$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-96}{2*48}=\frac{-192}{96} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+96}{2*48}=\frac{0}{96} =0 $
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