(5-z)(3z-7)=0

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Solution for (5-z)(3z-7)=0 equation:



(5-z)(3z-7)=0
We add all the numbers together, and all the variables
(-1z+5)(3z-7)=0
We multiply parentheses ..
(-3z^2+7z+15z-35)=0
We get rid of parentheses
-3z^2+7z+15z-35=0
We add all the numbers together, and all the variables
-3z^2+22z-35=0
a = -3; b = 22; c = -35;
Δ = b2-4ac
Δ = 222-4·(-3)·(-35)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-8}{2*-3}=\frac{-30}{-6} =+5 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*-3}=\frac{-14}{-6} =2+1/3 $

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