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4t^2-23t-6=0
a = 4; b = -23; c = -6;
Δ = b2-4ac
Δ = -232-4·4·(-6)
Δ = 625
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{625}=25$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*4}=\frac{-2}{8} =-1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*4}=\frac{48}{8} =6 $
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