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-2p^2=16p=24
We move all terms to the left:
-2p^2-(16p)=0
a = -2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-2)·0
Δ = 256
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{256}=16$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-2}=\frac{0}{-4} =0 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-2}=\frac{32}{-4} =-8 $
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