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x^2+10x-14000=0
a = 1; b = 10; c = -14000;
Δ = b2-4ac
Δ = 102-4·1·(-14000)
Δ = 56100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56100}=\sqrt{100*561}=\sqrt{100}*\sqrt{561}=10\sqrt{561}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{561}}{2*1}=\frac{-10-10\sqrt{561}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{561}}{2*1}=\frac{-10+10\sqrt{561}}{2} $
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