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3r^2-2r+15r=10
We move all terms to the left:
3r^2-2r+15r-(10)=0
We add all the numbers together, and all the variables
3r^2+13r-10=0
a = 3; b = 13; c = -10;
Δ = b2-4ac
Δ = 132-4·3·(-10)
Δ = 289
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{289}=17$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*3}=\frac{-30}{6} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*3}=\frac{4}{6} =2/3 $
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