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16q^2=100
We move all terms to the left:
16q^2-(100)=0
a = 16; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·16·(-100)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{6400}=80$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*16}=\frac{-80}{32} =-2+1/2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*16}=\frac{80}{32} =2+1/2 $
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