(4x^2)-16x+12=0

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Solution for (4x^2)-16x+12=0 equation: 



(4x^2)-16x+12=0

a = 4; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·4·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*4}=\frac{8}{8}
=1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*4}=\frac{24}{8}
=3
$

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