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2z(5z+2)=-15z-6
We move all terms to the left:
2z(5z+2)-(-15z-6)=0
We multiply parentheses
10z^2+4z-(-15z-6)=0
We get rid of parentheses
10z^2+4z+15z+6=0
We add all the numbers together, and all the variables
10z^2+19z+6=0
a = 10; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·10·6
Δ = 121
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{121}=11$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*10}=\frac{-30}{20} =-1+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*10}=\frac{-8}{20} =-2/5 $
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