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

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Solution for (5+z)(4z+7)=0 equation:



(5+z)(4z+7)=0
We add all the numbers together, and all the variables
(z+5)(4z+7)=0
We multiply parentheses ..
(+4z^2+7z+20z+35)=0
We get rid of parentheses
4z^2+7z+20z+35=0
We add all the numbers together, and all the variables
4z^2+27z+35=0
a = 4; b = 27; c = +35;
Δ = b2-4ac
Δ = 272-4·4·35
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-13}{2*4}=\frac{-40}{8} =-5 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+13}{2*4}=\frac{-14}{8} =-1+3/4 $

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