(-7/2z+4)+(1/5z=15)=

Solution for (-7/2z+4)+(1/5z=15)= equation:

(-7/2z+4)+(1/5z=15)=

We move all terms to the left:

(-7/2z+4)+(1/5z-(15))=0


Domain of the equation: 2z+4)!=0

z∈R



Domain of the equation: 5z-15)!=0

z∈R


We get rid of parentheses

-7/2z+1/5z+4-15=0

We calculate fractions


(-35z)/10z^2+2z/10z^2+4-15=0

We add all the numbers together, and all the variables

(-35z)/10z^2+2z/10z^2-11=0

We multiply all the terms by the denominator


(-35z)+2z-11*10z^2=0

We add all the numbers together, and all the variables

2z+(-35z)-11*10z^2=0

Wy multiply elements

-110z^2+2z+(-35z)=0

We get rid of parentheses

-110z^2+2z-35z=0

We add all the numbers together, and all the variables

-110z^2-33z=0

a = -110; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·(-110)·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-33}{2*-110}=\frac{0}{-220}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*-110}=\frac{66}{-220}
=-3/10
$