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

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



(7-z)(4z+5)=0
We add all the numbers together, and all the variables
(-1z+7)(4z+5)=0
We multiply parentheses ..
(-4z^2-5z+28z+35)=0
We get rid of parentheses
-4z^2-5z+28z+35=0
We add all the numbers together, and all the variables
-4z^2+23z+35=0
a = -4; b = 23; c = +35;
Δ = b2-4ac
Δ = 232-4·(-4)·35
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*-4}=\frac{-56}{-8} =+7 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*-4}=\frac{10}{-8} =-1+1/4 $

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