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2x^2+50x-399=0
a = 2; b = 50; c = -399;
Δ = b2-4ac
Δ = 502-4·2·(-399)
Δ = 5692
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5692}=\sqrt{4*1423}=\sqrt{4}*\sqrt{1423}=2\sqrt{1423}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{1423}}{2*2}=\frac{-50-2\sqrt{1423}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{1423}}{2*2}=\frac{-50+2\sqrt{1423}}{4} $
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