10(z+3)=4(2z+9)4z

Solution for 10(z+3)=4(2z+9)4z equation:

10(z+3)=4(2z+9)4z

We move all terms to the left:

10(z+3)-(4(2z+9)4z)=0

We multiply parentheses

10z-(4(2z+9)4z)+30=0


We calculate terms in parentheses: -(4(2z+9)4z), so:

4(2z+9)4z

We multiply parentheses

32z^2+144z

Back to the equation:

-(32z^2+144z)


We get rid of parentheses

-32z^2+10z-144z+30=0

We add all the numbers together, and all the variables

-32z^2-134z+30=0

a = -32; b = -134; c = +30;
Δ = b2-4ac
Δ = -1342-4·(-32)·30
Δ = 21796
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:

$\sqrt{\Delta}=\sqrt{21796}=\sqrt{4*5449}=\sqrt{4}*\sqrt{5449}=2\sqrt{5449}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-134)-2\sqrt{5449}}{2*-32}=\frac{134-2\sqrt{5449}}{-64}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-134)+2\sqrt{5449}}{2*-32}=\frac{134+2\sqrt{5449}}{-64}
$