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6r^2+11r-19=5
We move all terms to the left:
6r^2+11r-19-(5)=0
We add all the numbers together, and all the variables
6r^2+11r-24=0
a = 6; b = 11; c = -24;
Δ = b2-4ac
Δ = 112-4·6·(-24)
Δ = 697
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{-(11)-\sqrt{697}}{2*6}=\frac{-11-\sqrt{697}}{12} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{697}}{2*6}=\frac{-11+\sqrt{697}}{12} $
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