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3x^2-3x-126=0
a = 3; b = -3; c = -126;
Δ = b2-4ac
Δ = -32-4·3·(-126)
Δ = 1521
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-39}{2*3}=\frac{-36}{6} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+39}{2*3}=\frac{42}{6} =7 $
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