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

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



(4z-5)(7+z)=0
We add all the numbers together, and all the variables
(4z-5)(z+7)=0
We multiply parentheses ..
(+4z^2+28z-5z-35)=0
We get rid of parentheses
4z^2+28z-5z-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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