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3z(5z+4)=5z+4.
We move all terms to the left:
3z(5z+4)-(5z+4.)=0
We add all the numbers together, and all the variables
3z(5z+4)-(5z+4)=0
We multiply parentheses
15z^2+12z-(5z+4)=0
We get rid of parentheses
15z^2+12z-5z-4=0
We add all the numbers together, and all the variables
15z^2+7z-4=0
a = 15; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·15·(-4)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*15}=\frac{-24}{30} =-4/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*15}=\frac{10}{30} =1/3 $
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