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4t^2-11t+5=0
a = 4; b = -11; c = +5;
Δ = b2-4ac
Δ = -112-4·4·5
Δ = 41
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{41}}{2*4}=\frac{11-\sqrt{41}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{41}}{2*4}=\frac{11+\sqrt{41}}{8} $
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