3s^2-16s+13=0

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Solution for 3s^2-16s+13=0 equation: 



3s^2-16s+13=0

a = 3; b = -16; c = +13;
Δ = b2-4ac
Δ = -162-4·3·13
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-10}{2*3}=\frac{6}{6}
=1
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+10}{2*3}=\frac{26}{6}
=4+1/3
$

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