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4t^2-23t+25=0
a = 4; b = -23; c = +25;
Δ = b2-4ac
Δ = -232-4·4·25
Δ = 129
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{-(-23)-\sqrt{129}}{2*4}=\frac{23-\sqrt{129}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{129}}{2*4}=\frac{23+\sqrt{129}}{8} $
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