-3v^2-22v+45=0

Solution for -3v^2-22v+45=0 equation:

-3v^2-22v+45=0

a = -3; b = -22; c = +45;
Δ = b2-4ac
Δ = -222-4·(-3)·45
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-32}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+32}{2*-3}=\frac{54}{-6}
=-9
$