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3x^2+24x-220=0
a = 3; b = 24; c = -220;
Δ = b2-4ac
Δ = 242-4·3·(-220)
Δ = 3216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3216}=\sqrt{16*201}=\sqrt{16}*\sqrt{201}=4\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{201}}{2*3}=\frac{-24-4\sqrt{201}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{201}}{2*3}=\frac{-24+4\sqrt{201}}{6} $
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