1/5(25z-30)=2/4(12z+16)

Solution for 1/5(25z-30)=2/4(12z+16) equation:

1/5(25z-30)=2/4(12z+16)

We move all terms to the left:

1/5(25z-30)-(2/4(12z+16))=0


Domain of the equation: 5(25z-30)!=0

z∈R



Domain of the equation: 4(12z+16))!=0

z∈R


We calculate fractions


(4z1/(5(25z-30)*4(12z+16)))+(-10z2/(5(25z-30)*4(12z+16)))=0


We calculate terms in parentheses: +(4z1/(5(25z-30)*4(12z+16))), so:

4z1/(5(25z-30)*4(12z+16))

We multiply all the terms by the denominator


4z1

We add all the numbers together, and all the variables

4z

Back to the equation:

+(4z)



We calculate terms in parentheses: +(-10z2/(5(25z-30)*4(12z+16))), so:

-10z2/(5(25z-30)*4(12z+16))

We multiply all the terms by the denominator


-10z2

We add all the numbers together, and all the variables

-10z^2

Back to the equation:

+(-10z^2)


We get rid of parentheses

-10z^2+4z=0

a = -10; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-10)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-10}=\frac{-8}{-20}
=2/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-10}=\frac{0}{-20}
=0
$