Solve (x-4)^2+(x+9)^2=2x^2

$$( x - 4 ) ^ { 2 } + ( x + 9 ) ^ { 2 } = 2 x ^ { 2 }$$

Solve for x

$$x = -\frac{97}{10} = -9\frac{7}{10} = -9.7$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16+\left(x+9\right)^{2}=2x^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+9\right)^{2}.

$$x^{2}-8x+16+x^{2}+18x+81=2x^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-8x+16+18x+81=2x^{2}$$

Combine -8x and 18x to get 10x.

$$2x^{2}+10x+16+81=2x^{2}$$

Add 16 and 81 to get 97.

$$2x^{2}+10x+97=2x^{2}$$

Subtract 2x^{2} from both sides.

$$2x^{2}+10x+97-2x^{2}=0$$

Combine 2x^{2} and -2x^{2} to get 0.

$$10x+97=0$$

Subtract 97 from both sides. Anything subtracted from zero gives its negation.

$$10x=-97$$

Divide both sides by 10.

$$x=\frac{-97}{10}$$

Fraction \frac{-97}{10} can be rewritten as -\frac{97}{10} by extracting the negative sign.

$$x=-\frac{97}{10}$$