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1.5x^2+x-4.6=0
a = 1.5; b = 1; c = -4.6;
Δ = b2-4ac
Δ = 12-4·1.5·(-4.6)
Δ = 28.6
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{28.6}}{2*1.5}=\frac{-1-\sqrt{28.6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{28.6}}{2*1.5}=\frac{-1+\sqrt{28.6}}{3} $
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