-3u^2+5u+2=0

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Solution for -3u^2+5u+2=0 equation: 



-3u^2+5u+2=0

a = -3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·(-3)·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-3}=\frac{-12}{-6}
=+2
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-3}=\frac{2}{-6}
=-1/3
$

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