2t^2+14t-36=0

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Solution for 2t^2+14t-36=0 equation: 



2t^2+14t-36=0

a = 2; b = 14; c = -36;
Δ = b2-4ac
Δ = 142-4·2·(-36)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-22}{2*2}=\frac{-36}{4}
=-9
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+22}{2*2}=\frac{8}{4}
=2
$

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