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3t^2-4t=7
We move all terms to the left:
3t^2-4t-(7)=0
a = 3; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·3·(-7)
Δ = 100
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{100}=10$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*3}=\frac{-6}{6} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*3}=\frac{14}{6} =2+1/3 $
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