Solve 9x^2+1+16=(3x+4)^2

$$9 x ^ { 2 } + 1 + 16 = ( 3 x + 4 ) ^ { 2 }$$

Solve for x

$$x=\frac{1}{24}\approx 0.041666667$$

Steps for Solving Linear Equation

Add 1 and 16 to get 17.

$$9x^{2}+17=\left(3x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$9x^{2}+17=9x^{2}+24x+16$$

Subtract 9x^{2} from both sides.

$$9x^{2}+17-9x^{2}=24x+16$$

Combine 9x^{2} and -9x^{2} to get 0.

$$17=24x+16$$

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

$$24x+16=17$$

Subtract 16 from both sides.

$$24x=17-16$$

Subtract 16 from 17 to get 1.

$$24x=1$$

Divide both sides by 24.

$$x=\frac{1}{24}$$