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2x^2+50x-6000=0
a = 2; b = 50; c = -6000;
Δ = b2-4ac
Δ = 502-4·2·(-6000)
Δ = 50500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{50500}=\sqrt{100*505}=\sqrt{100}*\sqrt{505}=10\sqrt{505}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{505}}{2*2}=\frac{-50-10\sqrt{505}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{505}}{2*2}=\frac{-50+10\sqrt{505}}{4} $
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