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3t^2+9=63
We move all terms to the left:
3t^2+9-(63)=0
We add all the numbers together, and all the variables
3t^2-54=0
a = 3; b = 0; c = -54;
Δ = b2-4ac
Δ = 02-4·3·(-54)
Δ = 648
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*3}=\frac{0-18\sqrt{2}}{6} =-\frac{18\sqrt{2}}{6} =-3\sqrt{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*3}=\frac{0+18\sqrt{2}}{6} =\frac{18\sqrt{2}}{6} =3\sqrt{2} $
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