4z^2+7z+4=6

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Solution for 4z^2+7z+4=6 equation:



4z^2+7z+4=6
We move all terms to the left:
4z^2+7z+4-(6)=0
We add all the numbers together, and all the variables
4z^2+7z-2=0
a = 4; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·4·(-2)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*4}=\frac{-16}{8} =-2 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*4}=\frac{2}{8} =1/4 $

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