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3x^2+100x-10000=0
a = 3; b = 100; c = -10000;
Δ = b2-4ac
Δ = 1002-4·3·(-10000)
Δ = 130000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{130000}=\sqrt{10000*13}=\sqrt{10000}*\sqrt{13}=100\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-100\sqrt{13}}{2*3}=\frac{-100-100\sqrt{13}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+100\sqrt{13}}{2*3}=\frac{-100+100\sqrt{13}}{6} $
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