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126=3x^2+3x
We move all terms to the left:
126-(3x^2+3x)=0
We get rid of parentheses
-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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