-3x^2+29x-18=0

Solution for -3x^2+29x-18=0 equation:

-3x^2+29x-18=0

a = -3; b = 29; c = -18;
Δ = b2-4ac
Δ = 292-4·(-3)·(-18)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-25}{2*-3}=\frac{-54}{-6}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+25}{2*-3}=\frac{-4}{-6}
=2/3
$