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1.5x^2+44x+120=0
a = 1.5; b = 44; c = +120;
Δ = b2-4ac
Δ = 442-4·1.5·120
Δ = 1216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-8\sqrt{19}}{2*1.5}=\frac{-44-8\sqrt{19}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+8\sqrt{19}}{2*1.5}=\frac{-44+8\sqrt{19}}{3} $
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