Let's solve in <math><mi>ℝ</mi></math>: <math><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo> + </mo><mn>2</mn><mi>x</mi><mo> + </mo><mn>1</mn><mo> = </mo><mn>0</mn></math>
You can rephrase the expression into: <math><msup><mrow><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo></mrow><mrow><mn>2</mn></mrow></msup></math>
<math><msup><mrow><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo> = </mo><mn>0</mn></math> is equivalent to: <math><mi>x</mi><mo> + </mo><mn>1</mn><mo> = </mo><mn>0</mn></math>

▷<b>Result: &nbsp; <math><mi>x</mi><mo> = </mo><mn>-1</mn></math></b><span style='font-size:smaller'> &nbsp; &nbsp; (processing: 3 steps, 2 ms)</span>