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1.5x^2+4x-299=0
a = 1.5; b = 4; c = -299;
Δ = b2-4ac
Δ = 42-4·1.5·(-299)
Δ = 1810
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{-(4)-\sqrt{1810}}{2*1.5}=\frac{-4-\sqrt{1810}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{1810}}{2*1.5}=\frac{-4+\sqrt{1810}}{3} $
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