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16b^2+52b+36=
We move all terms to the left:
16b^2+52b+36-()=0
We add all the numbers together, and all the variables
16b^2+52b=0
a = 16; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·16·0
Δ = 2704
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{2704}=52$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*16}=\frac{-104}{32} =-3+1/4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*16}=\frac{0}{32} =0 $
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