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0.3x^2-4x-2.5=0
a = 0.3; b = -4; c = -2.5;
Δ = b2-4ac
Δ = -42-4·0.3·(-2.5)
Δ = 19
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{19}}{2*0.3}=\frac{4-\sqrt{19}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{19}}{2*0.3}=\frac{4+\sqrt{19}}{0.6} $
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