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(5z+1)(4z+3)=170
We move all terms to the left:
(5z+1)(4z+3)-(170)=0
We multiply parentheses ..
(+20z^2+15z+4z+3)-170=0
We get rid of parentheses
20z^2+15z+4z+3-170=0
We add all the numbers together, and all the variables
20z^2+19z-167=0
a = 20; b = 19; c = -167;
Δ = b2-4ac
Δ = 192-4·20·(-167)
Δ = 13721
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{13721}}{2*20}=\frac{-19-\sqrt{13721}}{40} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{13721}}{2*20}=\frac{-19+\sqrt{13721}}{40} $
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