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3x^2+12x-160=0
a = 3; b = 12; c = -160;
Δ = b2-4ac
Δ = 122-4·3·(-160)
Δ = 2064
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{129}}{2*3}=\frac{-12-4\sqrt{129}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{129}}{2*3}=\frac{-12+4\sqrt{129}}{6} $
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