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4x^2-16x=-8
We move all terms to the left:
4x^2-16x-(-8)=0
We add all the numbers together, and all the variables
4x^2-16x+8=0
a = 4; b = -16; c = +8;
Δ = b2-4ac
Δ = -162-4·4·8
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{2}}{2*4}=\frac{16-8\sqrt{2}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*4}=\frac{16+8\sqrt{2}}{8} $
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