t^2+6t-72=0

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Solution for t^2+6t-72=0 equation:



t^2+6t-72=0
a = 1; b = 6; c = -72;
Δ = b2-4ac
Δ = 62-4·1·(-72)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*1}=\frac{-24}{2} =-12 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*1}=\frac{12}{2} =6 $

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