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20-2x^2=4x
We move all terms to the left:
20-2x^2-(4x)=0
a = -2; b = -4; c = +20;
Δ = b2-4ac
Δ = -42-4·(-2)·20
Δ = 176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*-2}=\frac{4-4\sqrt{11}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*-2}=\frac{4+4\sqrt{11}}{-4} $
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