Solve x=1.1left(6-yright)^2-3

$$x = 1.1 { \left(6-y \right) }^{ 2 } -3$$

Solve for x

$$x=\frac{11\left(6-y\right)^{2}}{10}-3$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$x=1.1\left(36-12y+y^{2}\right)-3$$

Use the distributive property to multiply 1.1 by 36-12y+y^{2}.

$$x=39.6-13.2y+1.1y^{2}-3$$

Subtract 3 from 39.6 to get 36.6.

$$x=36.6-13.2y+1.1y^{2}$$

Solve for y (complex solution)

$$y=\sqrt{\frac{10x+30}{11}}+6$$

$$y=-\sqrt{\frac{10x+30}{11}}+6$$

Solve for y

$$y=\frac{\sqrt{\frac{40x+120}{11}}}{2}+6$$

$$y=-\frac{\sqrt{\frac{40x+120}{11}}}{2}+6\text{, }x\geq -3$$