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3x^2+12x-0.23=0
a = 3; b = 12; c = -0.23;
Δ = b2-4ac
Δ = 122-4·3·(-0.23)
Δ = 146.76
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{-(12)-\sqrt{146.76}}{2*3}=\frac{-12-\sqrt{146.76}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{146.76}}{2*3}=\frac{-12+\sqrt{146.76}}{6} $
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