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(3/5)z-5=10
We move all terms to the left:
(3/5)z-5-(10)=0
Domain of the equation: 5)z!=0We add all the numbers together, and all the variables
z!=0/1
z!=0
z∈R
(+3/5)z-5-10=0
We add all the numbers together, and all the variables
(+3/5)z-15=0
We multiply parentheses
3z^2-15=0
a = 3; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·3·(-15)
Δ = 180
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*3}=\frac{0-6\sqrt{5}}{6} =-\frac{6\sqrt{5}}{6} =-\sqrt{5} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*3}=\frac{0+6\sqrt{5}}{6} =\frac{6\sqrt{5}}{6} =\sqrt{5} $
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