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2x^2+2300x+600000=0
a = 2; b = 2300; c = +600000;
Δ = b2-4ac
Δ = 23002-4·2·600000
Δ = 490000
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{490000}=700$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2300)-700}{2*2}=\frac{-3000}{4} =-750 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2300)+700}{2*2}=\frac{-1600}{4} =-400 $
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