Solve 1=left(2.1xright)^2+x^2

$$1= { \left(2.1x \right) }^{ 2 } + { x }^{ 2 }$$

Solve for x

$$x=\frac{10\sqrt{541}}{541}\approx 0.42993358$$

$$x=-\frac{10\sqrt{541}}{541}\approx -0.42993358$$

Steps by Finding Square Root

Expand \left(2.1x\right)^{2}.

$$1=2.1^{2}x^{2}+x^{2}$$

Calculate 2.1 to the power of 2 and get 4.41.

$$1=4.41x^{2}+x^{2}$$

Combine 4.41x^{2} and x^{2} to get 5.41x^{2}.

$$1=5.41x^{2}$$

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

$$5.41x^{2}=1$$

Divide both sides by 5.41.

$$x^{2}=\frac{1}{5.41}$$

Expand \frac{1}{5.41} by multiplying both numerator and the denominator by 100.

$$x^{2}=\frac{100}{541}$$

Take the square root of both sides of the equation.

$$x=\frac{10\sqrt{541}}{541} x=-\frac{10\sqrt{541}}{541}$$

Steps Using the Quadratic Formula

Expand \left(2.1x\right)^{2}.

$$1=2.1^{2}x^{2}+x^{2}$$

Calculate 2.1 to the power of 2 and get 4.41.

$$1=4.41x^{2}+x^{2}$$

Combine 4.41x^{2} and x^{2} to get 5.41x^{2}.

$$1=5.41x^{2}$$

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

$$5.41x^{2}=1$$

Subtract 1 from both sides.

$$5.41x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5.41 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5.41\left(-1\right)}}{2\times 5.41}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5.41\left(-1\right)}}{2\times 5.41}$$

Multiply -4 times 5.41.

$$x=\frac{0±\sqrt{-21.64\left(-1\right)}}{2\times 5.41}$$

Multiply -21.64 times -1.

$$x=\frac{0±\sqrt{21.64}}{2\times 5.41}$$

Take the square root of 21.64.

$$x=\frac{0±\frac{\sqrt{541}}{5}}{2\times 5.41}$$

Multiply 2 times 5.41.

$$x=\frac{0±\frac{\sqrt{541}}{5}}{10.82}$$

Now solve the equation x=\frac{0±\frac{\sqrt{541}}{5}}{10.82} when ± is plus.

$$x=\frac{10\sqrt{541}}{541}$$

Now solve the equation x=\frac{0±\frac{\sqrt{541}}{5}}{10.82} when ± is minus.

$$x=-\frac{10\sqrt{541}}{541}$$

The equation is now solved.

$$x=\frac{10\sqrt{541}}{541} x=-\frac{10\sqrt{541}}{541}$$