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(4-z)(3z+5)=0
We add all the numbers together, and all the variables
(-1z+4)(3z+5)=0
We multiply parentheses ..
(-3z^2-5z+12z+20)=0
We get rid of parentheses
-3z^2-5z+12z+20=0
We add all the numbers together, and all the variables
-3z^2+7z+20=0
a = -3; b = 7; c = +20;
Δ = b2-4ac
Δ = 72-4·(-3)·20
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*-3}=\frac{-24}{-6} =+4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*-3}=\frac{10}{-6} =-1+2/3 $
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