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1.5x^2+3x-20=0
a = 1.5; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·1.5·(-20)
Δ = 129
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{129}}{2*1.5}=\frac{-3-\sqrt{129}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*1.5}=\frac{-3+\sqrt{129}}{3} $
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