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-1.5x^2+5x+20=0
a = -1.5; b = 5; c = +20;
Δ = b2-4ac
Δ = 52-4·(-1.5)·20
Δ = 145
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{-(5)-\sqrt{145}}{2*-1.5}=\frac{-5-\sqrt{145}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{145}}{2*-1.5}=\frac{-5+\sqrt{145}}{-3} $
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