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(2z+3)(8+z)=0
We add all the numbers together, and all the variables
(2z+3)(z+8)=0
We multiply parentheses ..
(+2z^2+16z+3z+24)=0
We get rid of parentheses
2z^2+16z+3z+24=0
We add all the numbers together, and all the variables
2z^2+19z+24=0
a = 2; b = 19; c = +24;
Δ = b2-4ac
Δ = 192-4·2·24
Δ = 169
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{169}=13$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*2}=\frac{-32}{4} =-8 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*2}=\frac{-6}{4} =-1+1/2 $
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