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0.2x^2+10x-1500=1
We move all terms to the left:
0.2x^2+10x-1500-(1)=0
We add all the numbers together, and all the variables
0.2x^2+10x-1501=0
a = 0.2; b = 10; c = -1501;
Δ = b2-4ac
Δ = 102-4·0.2·(-1501)
Δ = 1300.8
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{-(10)-\sqrt{1300.8}}{2*0.2}=\frac{-10-\sqrt{1300.8}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{1300.8}}{2*0.2}=\frac{-10+\sqrt{1300.8}}{0.4} $
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