Solve (1/2y+13)*(y+13)=z+598

$$( \frac { 1 } { 2 } y + 13 ) \times ( y + 13 ) = z + 598$$

Solve for z

$$z=\frac{y^{2}+39y-858}{2}$$

Solution Steps

Use the distributive property to multiply \frac{1}{2}y+13 by y+13 and combine like terms.

$$\frac{1}{2}y^{2}+\frac{39}{2}y+169=z+598$$

Swap sides so that all variable terms are on the left hand side.

$$z+598=\frac{1}{2}y^{2}+\frac{39}{2}y+169$$

Subtract 598 from both sides.

$$z=\frac{1}{2}y^{2}+\frac{39}{2}y+169-598$$

Subtract 598 from 169 to get -429.

$$z=\frac{1}{2}y^{2}+\frac{39}{2}y-429$$

Solve for y

$$y=\frac{\sqrt{8z+4953}-39}{2}$$

$$y=\frac{-\sqrt{8z+4953}-39}{2}\text{, }z\geq -\frac{4953}{8}$$