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5z(2z+3)=4z+6
We move all terms to the left:
5z(2z+3)-(4z+6)=0
We multiply parentheses
10z^2+15z-(4z+6)=0
We get rid of parentheses
10z^2+15z-4z-6=0
We add all the numbers together, and all the variables
10z^2+11z-6=0
a = 10; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·10·(-6)
Δ = 361
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{361}=19$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*10}=\frac{-30}{20} =-1+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*10}=\frac{8}{20} =2/5 $
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