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x.20(2x-8)=440
We move all terms to the left:
x.20(2x-8)-(440)=0
We multiply parentheses
2x^2-8x-440=0
a = 2; b = -8; c = -440;
Δ = b2-4ac
Δ = -82-4·2·(-440)
Δ = 3584
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16\sqrt{14}}{2*2}=\frac{8-16\sqrt{14}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16\sqrt{14}}{2*2}=\frac{8+16\sqrt{14}}{4} $
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