(2z+1)(8-z)=0

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Solution for (2z+1)(8-z)=0 equation:



(2z+1)(8-z)=0
We add all the numbers together, and all the variables
(2z+1)(-1z+8)=0
We multiply parentheses ..
(-2z^2+16z-1z+8)=0
We get rid of parentheses
-2z^2+16z-1z+8=0
We add all the numbers together, and all the variables
-2z^2+15z+8=0
a = -2; b = 15; c = +8;
Δ = b2-4ac
Δ = 152-4·(-2)·8
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-17}{2*-2}=\frac{-32}{-4} =+8 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+17}{2*-2}=\frac{2}{-4} =-1/2 $

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