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4t^2-21t+15.75=0
a = 4; b = -21; c = +15.75;
Δ = b2-4ac
Δ = -212-4·4·15.75
Δ = 189
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{21}}{2*4}=\frac{21-3\sqrt{21}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{21}}{2*4}=\frac{21+3\sqrt{21}}{8} $
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