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-16x^2+16x+4=0
a = -16; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·(-16)·4
Δ = 512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*-16}=\frac{-16-16\sqrt{2}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*-16}=\frac{-16+16\sqrt{2}}{-32} $
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