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2r^2+13r-7=0
a = 2; b = 13; c = -7;
Δ = b2-4ac
Δ = 132-4·2·(-7)
Δ = 225
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{225}=15$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-15}{2*2}=\frac{-28}{4} =-7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+15}{2*2}=\frac{2}{4} =1/2 $
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