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1.5x^2+3x=120
We move all terms to the left:
1.5x^2+3x-(120)=0
a = 1.5; b = 3; c = -120;
Δ = b2-4ac
Δ = 32-4·1.5·(-120)
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*1.5}=\frac{-30}{3} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*1.5}=\frac{24}{3} =8 $
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