0=(16t^2+72t-81)

Solution for 0=(16t^2+72t-81) equation:

0=(16t^2+72t-81)

We move all terms to the left:

0-((16t^2+72t-81))=0

We add all the numbers together, and all the variables

-((16t^2+72t-81))=0


We calculate terms in parentheses: -((16t^2+72t-81)), so:

(16t^2+72t-81)

We get rid of parentheses

16t^2+72t-81

Back to the equation:

-(16t^2+72t-81)


We get rid of parentheses

-16t^2-72t+81=0

a = -16; b = -72; c = +81;
Δ = b2-4ac
Δ = -722-4·(-16)·81
Δ = 10368
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{10368}=\sqrt{5184*2}=\sqrt{5184}*\sqrt{2}=72\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-72\sqrt{2}}{2*-16}=\frac{72-72\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+72\sqrt{2}}{2*-16}=\frac{72+72\sqrt{2}}{-32}
$