2s^2-12s+16=0

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Solution for 2s^2-12s+16=0 equation: 



2s^2-12s+16=0

a = 2; b = -12; c = +16;
Δ = b2-4ac
Δ = -122-4·2·16
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*2}=\frac{8}{4}
=2
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*2}=\frac{16}{4}
=4
$

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