-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!=0

z!=0/2

z!=0

z∈R



Domain of the equation: 5z!=0

z!=0/5

z!=0

z∈R


We add all the numbers together, and all the variables

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

We calculate fractions


(-35z)/10z^2+(1*2z)/10z^2+4=0

We add all the numbers together, and all the variables

(-35z)/10z^2+(+1*2z)/10z^2+4=0

We multiply all the terms by the denominator


(-35z)+(+1*2z)+4*10z^2=0

Wy multiply elements

40z^2+(-35z)+(+1*2z)=0

We get rid of parentheses

40z^2-35z+1*2z=0

Wy multiply elements

40z^2-35z+2z=0

We add all the numbers together, and all the variables

40z^2-33z=0

a = 40; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·40·0
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*40}=\frac{0}{80}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*40}=\frac{66}{80}
=33/40
$