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.3x^2+.8x-2=0
a = .3; b = .8; c = -2;
Δ = b2-4ac
Δ = .82-4·.3·(-2)
Δ = 3.04
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{-(.8)-\sqrt{3.04}}{2*.3}=\frac{-0.8-\sqrt{3.04}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.8)+\sqrt{3.04}}{2*.3}=\frac{-0.8+\sqrt{3.04}}{0.6} $
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