t^2-15t-100=0

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Solution for t^2-15t-100=0 equation:



t^2-15t-100=0
a = 1; b = -15; c = -100;
Δ = b2-4ac
Δ = -152-4·1·(-100)
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{-(-15)-25}{2*1}=\frac{-10}{2} =-5 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*1}=\frac{40}{2} =20 $

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