Solve for z
z=2\left(x+3\right)^{2}-5
Solve for x
x=-\frac{\sqrt{2z+10}}{2}-3
x=\frac{\sqrt{2z+10}}{2}-3\text{, }z\geq -5
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z=2\left(x^{2}+6x+9\right)-5
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.
z=2x^{2}+12x+18-5
Use the distributive property to multiply 2 by x^{2}+6x+9.
z=2x^{2}+12x+13
Subtract 5 from 18 to get 13.
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