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0.3x^2+6.8x-125.8=0
a = 0.3; b = 6.8; c = -125.8;
Δ = b2-4ac
Δ = 6.82-4·0.3·(-125.8)
Δ = 197.2
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{-(6.8)-\sqrt{197.2}}{2*0.3}=\frac{-6.8-\sqrt{197.2}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6.8)+\sqrt{197.2}}{2*0.3}=\frac{-6.8+\sqrt{197.2}}{0.6} $
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