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2r^2+r-120=2r
We move all terms to the left:
2r^2+r-120-(2r)=0
We add all the numbers together, and all the variables
2r^2-1r-120=0
a = 2; b = -1; c = -120;
Δ = b2-4ac
Δ = -12-4·2·(-120)
Δ = 961
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{961}=31$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*2}=\frac{-30}{4} =-7+1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*2}=\frac{32}{4} =8 $
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