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2(x^2+16x-140)=0
We multiply parentheses
2x^2+32x-280=0
a = 2; b = 32; c = -280;
Δ = b2-4ac
Δ = 322-4·2·(-280)
Δ = 3264
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3264}=\sqrt{64*51}=\sqrt{64}*\sqrt{51}=8\sqrt{51}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{51}}{2*2}=\frac{-32-8\sqrt{51}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{51}}{2*2}=\frac{-32+8\sqrt{51}}{4} $
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