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0.45x^2+47x-1200=0
a = 0.45; b = 47; c = -1200;
Δ = b2-4ac
Δ = 472-4·0.45·(-1200)
Δ = 4369
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{-(47)-\sqrt{4369}}{2*0.45}=\frac{-47-\sqrt{4369}}{0.9} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+\sqrt{4369}}{2*0.45}=\frac{-47+\sqrt{4369}}{0.9} $
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