t^2-20t+96=0

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Solution for t^2-20t+96=0 equation: 



t^2-20t+96=0

a = 1; b = -20; c = +96;
Δ = b2-4ac
Δ = -202-4·1·96
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*1}=\frac{16}{2}
=8
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*1}=\frac{24}{2}
=12
$

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