-60z+4z=-7z(4+10z)

Solution for -60z+4z=-7z(4+10z) equation:

-60z+4z=-7z(4+10z)

We move all terms to the left:

-60z+4z-(-7z(4+10z))=0

We add all the numbers together, and all the variables

-60z+4z-(-7z(10z+4))=0

We add all the numbers together, and all the variables

-56z-(-7z(10z+4))=0


We calculate terms in parentheses: -(-7z(10z+4)), so:

-7z(10z+4)

We multiply parentheses

-70z^2-28z

Back to the equation:

-(-70z^2-28z)


We get rid of parentheses

70z^2+28z-56z=0

We add all the numbers together, and all the variables

70z^2-28z=0

a = 70; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·70·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*70}=\frac{0}{140}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*70}=\frac{56}{140}
=2/5
$