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1000x^2+16x-16=0
a = 1000; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·1000·(-16)
Δ = 64256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{64256}=\sqrt{256*251}=\sqrt{256}*\sqrt{251}=16\sqrt{251}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{251}}{2*1000}=\frac{-16-16\sqrt{251}}{2000} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{251}}{2*1000}=\frac{-16+16\sqrt{251}}{2000} $
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