1/5(z+3)^2=7

Solution for 1/5(z+3)^2=7 equation:

z in (-oo:+oo)

(1/5)*(z+3)^2 = 7 // - 7

(1/5)*(z+3)^2-7 = 0

1/5*(z+3)^2-7 = 0

1/5*(z+3)^2-7 = 0

1/5*z^2+6/5*z-26/5 = 0

1/5*z^2+6/5*z-26/5 = 0

1/5*(z^2+6*z-26) = 0

z^2+6*z-26 = 0

DELTA = 6^2-(-26*1*4)

DELTA = 140

DELTA > 0

z = (140^(1/2)-6)/(1*2) or z = (-140^(1/2)-6)/(1*2)

z = (2*35^(1/2)-6)/2 or z = (-2*35^(1/2)-6)/2

1/5*(z-((-2*35^(1/2)-6)/2))*(z-((2*35^(1/2)-6)/2)) = 0

1/5*(z-((-2*35^(1/2)-6)/2))*(z-((2*35^(1/2)-6)/2)) = 0

( z-((-2*35^(1/2)-6)/2) )

z-((-2*35^(1/2)-6)/2) = 0 // + (-2*35^(1/2)-6)/2

z = (-2*35^(1/2)-6)/2

( z-((2*35^(1/2)-6)/2) )

z-((2*35^(1/2)-6)/2) = 0 // + (2*35^(1/2)-6)/2

z = (2*35^(1/2)-6)/2

z in { (-2*35^(1/2)-6)/2, (2*35^(1/2)-6)/2 }