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8t^2-22t+5=0
a = 8; b = -22; c = +5;
Δ = b2-4ac
Δ = -222-4·8·5
Δ = 324
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{324}=18$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-18}{2*8}=\frac{4}{16} =1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+18}{2*8}=\frac{40}{16} =2+1/2 $
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