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(2z+3)(2+z)=0
We add all the numbers together, and all the variables
(2z+3)(z+2)=0
We multiply parentheses ..
(+2z^2+4z+3z+6)=0
We get rid of parentheses
2z^2+4z+3z+6=0
We add all the numbers together, and all the variables
2z^2+7z+6=0
a = 2; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·2·6
Δ = 1
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{1}=1$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*2}=\frac{-8}{4} =-2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*2}=\frac{-6}{4} =-1+1/2 $
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