3x^2+100x+400=0

Solution for 3x^2+100x+400=0 equation:

3x^2+100x+400=0

a = 3; b = 100; c = +400;
Δ = b2-4ac
Δ = 1002-4·3·400
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{13}}{2*3}=\frac{-100-20\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{13}}{2*3}=\frac{-100+20\sqrt{13}}{6}
$