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2x^2+5x-1500=0
a = 2; b = 5; c = -1500;
Δ = b2-4ac
Δ = 52-4·2·(-1500)
Δ = 12025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12025}=\sqrt{25*481}=\sqrt{25}*\sqrt{481}=5\sqrt{481}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{481}}{2*2}=\frac{-5-5\sqrt{481}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{481}}{2*2}=\frac{-5+5\sqrt{481}}{4} $
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