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0.2x^2+10x-1500=0
a = 0.2; b = 10; c = -1500;
Δ = b2-4ac
Δ = 102-4·0.2·(-1500)
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{13}}{2*0.2}=\frac{-10-10\sqrt{13}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{13}}{2*0.2}=\frac{-10+10\sqrt{13}}{0.4} $
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