3v^2-33v+90=0

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Solution for 3v^2-33v+90=0 equation: 



3v^2-33v+90=0

a = 3; b = -33; c = +90;
Δ = b2-4ac
Δ = -332-4·3·90
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3}{2*3}=\frac{30}{6}
=5
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3}{2*3}=\frac{36}{6}
=6
$

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