3x^2+60x-400=0

Solution for 3x^2+60x-400=0 equation:

3x^2+60x-400=0

a = 3; b = 60; c = -400;
Δ = b2-4ac
Δ = 602-4·3·(-400)
Δ = 8400
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:

$\sqrt{\Delta}=\sqrt{8400}=\sqrt{400*21}=\sqrt{400}*\sqrt{21}=20\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{21}}{2*3}=\frac{-60-20\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{21}}{2*3}=\frac{-60+20\sqrt{21}}{6}
$