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2(x^2+12x-80)=0
We multiply parentheses
2x^2+24x-160=0
a = 2; b = 24; c = -160;
Δ = b2-4ac
Δ = 242-4·2·(-160)
Δ = 1856
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{29}}{2*2}=\frac{-24-8\sqrt{29}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{29}}{2*2}=\frac{-24+8\sqrt{29}}{4} $
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