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2x^2-20x-2400=0
a = 2; b = -20; c = -2400;
Δ = b2-4ac
Δ = -202-4·2·(-2400)
Δ = 19600
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{19600}=140$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-140}{2*2}=\frac{-120}{4} =-30 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+140}{2*2}=\frac{160}{4} =40 $
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