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2x(3x+1)=60
We move all terms to the left:
2x(3x+1)-(60)=0
We multiply parentheses
6x^2+2x-60=0
a = 6; b = 2; c = -60;
Δ = b2-4ac
Δ = 22-4·6·(-60)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-38}{2*6}=\frac{-40}{12} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+38}{2*6}=\frac{36}{12} =3 $
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