z^2+3-4z=0

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Solution for z^2+3-4z=0 equation: 



z^2+3-4z=0

a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


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


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*1}=\frac{6}{2}
=3
$

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