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8r^2+13r=6
We move all terms to the left:
8r^2+13r-(6)=0
a = 8; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·8·(-6)
Δ = 361
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{361}=19$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*8}=\frac{-32}{16} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*8}=\frac{6}{16} =3/8 $
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