2t2-5t-7=0

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Solution for 2t2-5t-7=0 equation: 



2t^2-5t-7=0

a = 2; b = -5; c = -7;
Δ = b2-4ac
Δ = -52-4·2·(-7)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*2}=\frac{-4}{4}
=-1
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*2}=\frac{14}{4}
=3+1/2
$

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