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8r^2+13r-43=0
a = 8; b = 13; c = -43;
Δ = b2-4ac
Δ = 132-4·8·(-43)
Δ = 1545
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{1545}}{2*8}=\frac{-13-\sqrt{1545}}{16} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{1545}}{2*8}=\frac{-13+\sqrt{1545}}{16} $
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