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1.5x^2+3x-0.5=0
a = 1.5; b = 3; c = -0.5;
Δ = b2-4ac
Δ = 32-4·1.5·(-0.5)
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-2\sqrt{3}}{2*1.5}=\frac{-3-2\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+2\sqrt{3}}{2*1.5}=\frac{-3+2\sqrt{3}}{3} $
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