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16r^2-3=-2
We move all terms to the left:
16r^2-3-(-2)=0
We add all the numbers together, and all the variables
16r^2-1=0
a = 16; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·16·(-1)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*16}=\frac{-8}{32} =-1/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*16}=\frac{8}{32} =1/4 $
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