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.1x^2+3x-400=0
a = .1; b = 3; c = -400;
Δ = b2-4ac
Δ = 32-4·.1·(-400)
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*.1}=\frac{-16}{0.2} =-80 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*.1}=\frac{10}{0.2} =50 $
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