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1.5x^2+14x-39=0
a = 1.5; b = 14; c = -39;
Δ = b2-4ac
Δ = 142-4·1.5·(-39)
Δ = 430
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{-(14)-\sqrt{430}}{2*1.5}=\frac{-14-\sqrt{430}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+\sqrt{430}}{2*1.5}=\frac{-14+\sqrt{430}}{3} $
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