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3n^2-63=-12n
We move all terms to the left:
3n^2-63-(-12n)=0
We get rid of parentheses
3n^2+12n-63=0
a = 3; b = 12; c = -63;
Δ = b2-4ac
Δ = 122-4·3·(-63)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{900}=30$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-30}{2*3}=\frac{-42}{6} =-7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+30}{2*3}=\frac{18}{6} =3 $
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