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0.05x^2+12x-800=0
a = 0.05; b = 12; c = -800;
Δ = b2-4ac
Δ = 122-4·0.05·(-800)
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{19}}{2*0.05}=\frac{-12-4\sqrt{19}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{19}}{2*0.05}=\frac{-12+4\sqrt{19}}{0.1} $
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