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0.25r^2=49
We move all terms to the left:
0.25r^2-(49)=0
a = 0.25; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·0.25·(-49)
Δ = 49
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{49}=7$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-7}{2*0.25}=\frac{-7}{0.5} =-14 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+7}{2*0.25}=\frac{7}{0.5} =14 $
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