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3(x^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} $
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