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4z^2+3z-5=z
We move all terms to the left:
4z^2+3z-5-(z)=0
We add all the numbers together, and all the variables
4z^2+2z-5=0
a = 4; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·4·(-5)
Δ = 84
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*4}=\frac{-2-2\sqrt{21}}{8} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*4}=\frac{-2+2\sqrt{21}}{8} $
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