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5r(2r+6)=180
We move all terms to the left:
5r(2r+6)-(180)=0
We multiply parentheses
10r^2+30r-180=0
a = 10; b = 30; c = -180;
Δ = b2-4ac
Δ = 302-4·10·(-180)
Δ = 8100
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{8100}=90$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-90}{2*10}=\frac{-120}{20} =-6 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+90}{2*10}=\frac{60}{20} =3 $
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