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3x^2+3x=1260
We move all terms to the left:
3x^2+3x-(1260)=0
a = 3; b = 3; c = -1260;
Δ = b2-4ac
Δ = 32-4·3·(-1260)
Δ = 15129
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{15129}=123$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-123}{2*3}=\frac{-126}{6} =-21 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+123}{2*3}=\frac{120}{6} =20 $
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