12t^2-52t+40=0

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Solution for 12t^2-52t+40=0 equation: 



12t^2-52t+40=0

a = 12; b = -52; c = +40;
Δ = b2-4ac
Δ = -522-4·12·40
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-28}{2*12}=\frac{24}{24}
=1
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+28}{2*12}=\frac{80}{24}
=3+1/3
$

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