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

Solution for (-7/2z+4)+(1/5z+15)=0 equation:

(-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+19=0

We multiply all the terms by the denominator


(-35z)+2z+19*10z^2=0

We add all the numbers together, and all the variables

2z+(-35z)+19*10z^2=0

Wy multiply elements

190z^2+2z+(-35z)=0

We get rid of parentheses

190z^2+2z-35z=0

We add all the numbers together, and all the variables

190z^2-33z=0

a = 190; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·190·0
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*190}=\frac{0}{380}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*190}=\frac{66}{380}
=33/190
$