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1.5x^2+2.5x-1000=0
a = 1.5; b = 2.5; c = -1000;
Δ = b2-4ac
Δ = 2.52-4·1.5·(-1000)
Δ = 6006.25
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{-(2.5)-\sqrt{6006.25}}{2*1.5}=\frac{-2.5-\sqrt{6006.25}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{6006.25}}{2*1.5}=\frac{-2.5+\sqrt{6006.25}}{3} $
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