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.005x^2+.14x-35=0
a = .005; b = .14; c = -35;
Δ = b2-4ac
Δ = .142-4·.005·(-35)
Δ = 0.7196
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{-(.14)-\sqrt{0.7196}}{2*.005}=\frac{-0.14-\sqrt{0.7196}}{0.01} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.14)+\sqrt{0.7196}}{2*.005}=\frac{-0.14+\sqrt{0.7196}}{0.01} $
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