3x^2+4x-447=0

Solution for 3x^2+4x-447=0 equation:

3x^2+4x-447=0

a = 3; b = 4; c = -447;
Δ = b2-4ac
Δ = 42-4·3·(-447)
Δ = 5380
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:

$\sqrt{\Delta}=\sqrt{5380}=\sqrt{4*1345}=\sqrt{4}*\sqrt{1345}=2\sqrt{1345}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{1345}}{2*3}=\frac{-4-2\sqrt{1345}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{1345}}{2*3}=\frac{-4+2\sqrt{1345}}{6}
$