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.12x^2-2.4x+10=0
a = .12; b = -2.4; c = +10;
Δ = b2-4ac
Δ = -2.42-4·.12·10
Δ = 0.96
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{-(-2.4)-\sqrt{0.96}}{2*.12}=\frac{2.4-\sqrt{0.96}}{0.24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.4)+\sqrt{0.96}}{2*.12}=\frac{2.4+\sqrt{0.96}}{0.24} $
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