(7+z)(4z+1)=0

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Solution for (7+z)(4z+1)=0 equation:



(7+z)(4z+1)=0
We add all the numbers together, and all the variables
(z+7)(4z+1)=0
We multiply parentheses ..
(+4z^2+z+28z+7)=0
We get rid of parentheses
4z^2+z+28z+7=0
We add all the numbers together, and all the variables
4z^2+29z+7=0
a = 4; b = 29; c = +7;
Δ = b2-4ac
Δ = 292-4·4·7
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*4}=\frac{-56}{8} =-7 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*4}=\frac{-2}{8} =-1/4 $

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