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28r^2+16r-44=0
a = 28; b = 16; c = -44;
Δ = b2-4ac
Δ = 162-4·28·(-44)
Δ = 5184
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{5184}=72$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-72}{2*28}=\frac{-88}{56} =-1+4/7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+72}{2*28}=\frac{56}{56} =1 $
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