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(x+2)(2x)=140
We move all terms to the left:
(x+2)(2x)-(140)=0
We multiply parentheses
2x^2+4x-140=0
a = 2; b = 4; c = -140;
Δ = b2-4ac
Δ = 42-4·2·(-140)
Δ = 1136
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{71}}{2*2}=\frac{-4-4\sqrt{71}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{71}}{2*2}=\frac{-4+4\sqrt{71}}{4} $
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