-2t^2+5t+6=0

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Solution for -2t^2+5t+6=0 equation: 



-2t^2+5t+6=0

a = -2; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-2)·6
Δ = 73
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{-(5)-\sqrt{73}}{2*-2}=\frac{-5-\sqrt{73}}{-4}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*-2}=\frac{-5+\sqrt{73}}{-4}
$

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