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4r^2=-28r
We move all terms to the left:
4r^2-(-28r)=0
We get rid of parentheses
4r^2+28r=0
a = 4; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·4·0
Δ = 784
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{784}=28$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*4}=\frac{-56}{8} =-7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*4}=\frac{0}{8} =0 $
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