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-0.03x^2+3.7x-68=0
a = -0.03; b = 3.7; c = -68;
Δ = b2-4ac
Δ = 3.72-4·(-0.03)·(-68)
Δ = 5.53
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{-(3.7)-\sqrt{5.53}}{2*-0.03}=\frac{-3.7-\sqrt{5.53}}{-0.06} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.7)+\sqrt{5.53}}{2*-0.03}=\frac{-3.7+\sqrt{5.53}}{-0.06} $
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