Question

In the following exercises, solve by completing the square.

$5x^2-3x=-10$

Short answer

The solution is$x=\frac{3}{10}-\frac{\sqrt{191}i}{10},x=\frac{3}{10}+\frac{\sqrt{191}i}{10}$

Step-by-step solution

Step 1. Given information 

Equation:$5x^2-3x=-10$

Step 2. Calculation  

We have,

$5x^2-3x=-10$

$\frac{5x^2-3x}{5}=\frac{-10}{5}$

$x^2-\frac{3x}{5}+{\left(\frac{1}{2}·\left(-\frac{3}{5}\right)\right)}^2=-2+{\left(\frac{1}{2}·\left(-\frac{3}{5}\right)\right)}^2$

role="math" localid="1645182571593" $x^2-\frac{3x}{5}+\frac{9}{100}=-2+\frac{9}{100}$

role="math" localid="1645182680652" ${\left(x-\frac{3}{10}\right)}^2=-\frac{191}{100}$

role="math" localid="1645182688837" $\sqrt{{\left(x-\frac{3}{10}\right)}^2}=\sqrt{-\frac{191}{100}}$

$x-\frac{3}{10}=\pm \frac{\sqrt{191}i}{10}$

$x-\frac{3}{10}+\frac{3}{10}=\frac{3}{10}-\frac{\sqrt{191}i}{10},x-\frac{3}{10}+\frac{3}{10}=\frac{3}{10}+\frac{\sqrt{191}i}{10}$

$x=\frac{3}{10}-\frac{\sqrt{191}i}{10},x=\frac{3}{10}+\frac{\sqrt{191}i}{10}$