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1.5x^2-3x-5=0
a = 1.5; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·1.5·(-5)
Δ = 39
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{-(-3)-\sqrt{39}}{2*1.5}=\frac{3-\sqrt{39}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{39}}{2*1.5}=\frac{3+\sqrt{39}}{3} $
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