-3x^2+25x-42=0

Solution for -3x^2+25x-42=0 equation:

-3x^2+25x-42=0

a = -3; b = 25; c = -42;
Δ = b2-4ac
Δ = 252-4·(-3)·(-42)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-11}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+11}{2*-3}=\frac{-14}{-6}
=2+1/3
$