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40x+10x^2=17600
We move all terms to the left:
40x+10x^2-(17600)=0
a = 10; b = 40; c = -17600;
Δ = b2-4ac
Δ = 402-4·10·(-17600)
Δ = 705600
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{705600}=840$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-840}{2*10}=\frac{-880}{20} =-44 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+840}{2*10}=\frac{800}{20} =40 $
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