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3t^2-42t+144=0
a = 3; b = -42; c = +144;
Δ = b2-4ac
Δ = -422-4·3·144
Δ = 36
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{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6}{2*3}=\frac{36}{6} =6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6}{2*3}=\frac{48}{6} =8 $
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