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6(3t+12)=2t(3t+12)
We move all terms to the left:
6(3t+12)-(2t(3t+12))=0
We multiply parentheses
18t-(2t(3t+12))+72=0
We calculate terms in parentheses: -(2t(3t+12)), so:We get rid of parentheses
2t(3t+12)
We multiply parentheses
6t^2+24t
Back to the equation:
-(6t^2+24t)
-6t^2+18t-24t+72=0
We add all the numbers together, and all the variables
-6t^2-6t+72=0
a = -6; b = -6; c = +72;
Δ = b2-4ac
Δ = -62-4·(-6)·72
Δ = 1764
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-42}{2*-6}=\frac{-36}{-12} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+42}{2*-6}=\frac{48}{-12} =-4 $
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