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3x^2+8x-3500=0
a = 3; b = 8; c = -3500;
Δ = b2-4ac
Δ = 82-4·3·(-3500)
Δ = 42064
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{42064}=\sqrt{16*2629}=\sqrt{16}*\sqrt{2629}=4\sqrt{2629}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2629}}{2*3}=\frac{-8-4\sqrt{2629}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2629}}{2*3}=\frac{-8+4\sqrt{2629}}{6} $
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