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35r^2+16r-3=0
a = 35; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·35·(-3)
Δ = 676
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{676}=26$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-26}{2*35}=\frac{-42}{70} =-3/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+26}{2*35}=\frac{10}{70} =1/7 $
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