Solve y=10(x-3)^2+13

$$y = 10 ( x - 3 ) ^ { 2 } + 13$$

Solve for y

$$y=10\left(x-3\right)^{2}+13$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$y=10\left(x^{2}-6x+9\right)+13$$

Use the distributive property to multiply 10 by x^{2}-6x+9.

$$y=10x^{2}-60x+90+13$$

Add 90 and 13 to get 103.

$$y=10x^{2}-60x+103$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{10y-130}}{10}+3$$

$$x=\frac{\sqrt{10y-130}}{10}+3$$

Solve for x

$$x=-\frac{\sqrt{10y-130}}{10}+3$$

$$x=\frac{\sqrt{10y-130}}{10}+3\text{, }y\geq 13$$