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28=0.3r^2
We move all terms to the left:
28-(0.3r^2)=0
We get rid of parentheses
-0.3r^2+28=0
a = -0.3; b = 0; c = +28;
Δ = b2-4ac
Δ = 02-4·(-0.3)·28
Δ = 33.6
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{33.6}}{2*-0.3}=\frac{0-\sqrt{33.6}}{-0.6} =-\frac{\sqrt{}}{-0.6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{33.6}}{2*-0.3}=\frac{0+\sqrt{33.6}}{-0.6} =\frac{\sqrt{}}{-0.6} $
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