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4r^2-15=17r
We move all terms to the left:
4r^2-15-(17r)=0
a = 4; b = -17; c = -15;
Δ = b2-4ac
Δ = -172-4·4·(-15)
Δ = 529
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{529}=23$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-23}{2*4}=\frac{-6}{8} =-3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+23}{2*4}=\frac{40}{8} =5 $
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