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3x^2+14x-51=0
a = 3; b = 14; c = -51;
Δ = b2-4ac
Δ = 142-4·3·(-51)
Δ = 808
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{808}=\sqrt{4*202}=\sqrt{4}*\sqrt{202}=2\sqrt{202}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{202}}{2*3}=\frac{-14-2\sqrt{202}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{202}}{2*3}=\frac{-14+2\sqrt{202}}{6} $
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