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-0.3x^2+6x-16.7=0
a = -0.3; b = 6; c = -16.7;
Δ = b2-4ac
Δ = 62-4·(-0.3)·(-16.7)
Δ = 15.96
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{-(6)-\sqrt{15.96}}{2*-0.3}=\frac{-6-\sqrt{15.96}}{-0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+\sqrt{15.96}}{2*-0.3}=\frac{-6+\sqrt{15.96}}{-0.6} $
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