3x^2+45x-168=0

Solution for 3x^2+45x-168=0 equation:

3x^2+45x-168=0

a = 3; b = 45; c = -168;
Δ = b2-4ac
Δ = 452-4·3·(-168)
Δ = 4041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4041}=\sqrt{9*449}=\sqrt{9}*\sqrt{449}=3\sqrt{449}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-3\sqrt{449}}{2*3}=\frac{-45-3\sqrt{449}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+3\sqrt{449}}{2*3}=\frac{-45+3\sqrt{449}}{6}
$