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10x^2+20x-800=0
a = 10; b = 20; c = -800;
Δ = b2-4ac
Δ = 202-4·10·(-800)
Δ = 32400
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{32400}=180$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-180}{2*10}=\frac{-200}{20} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+180}{2*10}=\frac{160}{20} =8 $
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