3x^2+42x-400=0

Solution for 3x^2+42x-400=0 equation:

3x^2+42x-400=0

a = 3; b = 42; c = -400;
Δ = b2-4ac
Δ = 422-4·3·(-400)
Δ = 6564
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:

$\sqrt{\Delta}=\sqrt{6564}=\sqrt{4*1641}=\sqrt{4}*\sqrt{1641}=2\sqrt{1641}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{1641}}{2*3}=\frac{-42-2\sqrt{1641}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{1641}}{2*3}=\frac{-42+2\sqrt{1641}}{6}
$