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3t^2-22t+32=0
a = 3; b = -22; c = +32;
Δ = b2-4ac
Δ = -222-4·3·32
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10}{2*3}=\frac{12}{6} =2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10}{2*3}=\frac{32}{6} =5+1/3 $
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