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3t-4t^2=0
a = -4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-4)·0
Δ = 9
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{9}=3$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-4}=\frac{-6}{-8} =3/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-4}=\frac{0}{-8} =0 $
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