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2x(-2x+8)=4
We move all terms to the left:
2x(-2x+8)-(4)=0
We multiply parentheses
-4x^2+16x-4=0
a = -4; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·(-4)·(-4)
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{3}}{2*-4}=\frac{-16-8\sqrt{3}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{3}}{2*-4}=\frac{-16+8\sqrt{3}}{-8} $
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