Solve 10x^2-2x+2=3x-4

$$10 x ^ { 2 } - 2 x + 2 = 3 x - 4$$

Solve for x (complex solution)

$$x=\frac{\sqrt{215}i}{20}+\frac{1}{4}\approx 0.25+0.733143915i$$

$$x=-\frac{\sqrt{215}i}{20}+\frac{1}{4}\approx 0.25-0.733143915i$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$10x^{2}-2x+2-3x=-4$$

Combine -2x and -3x to get -5x.

$$10x^{2}-5x+2=-4$$

Add 4 to both sides.

$$10x^{2}-5x+2+4=0$$

Add 2 and 4 to get 6.

$$10x^{2}-5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 10\times 6}}{2\times 10}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 10\times 6}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-5\right)±\sqrt{25-40\times 6}}{2\times 10}$$

Multiply -40 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-240}}{2\times 10}$$

Add 25 to -240.

$$x=\frac{-\left(-5\right)±\sqrt{-215}}{2\times 10}$$

Take the square root of -215.

$$x=\frac{-\left(-5\right)±\sqrt{215}i}{2\times 10}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{215}i}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{5±\sqrt{215}i}{20}$$

Now solve the equation x=\frac{5±\sqrt{215}i}{20} when ± is plus. Add 5 to i\sqrt{215}.

$$x=\frac{5+\sqrt{215}i}{20}$$

Divide 5+i\sqrt{215} by 20.

$$x=\frac{\sqrt{215}i}{20}+\frac{1}{4}$$

Now solve the equation x=\frac{5±\sqrt{215}i}{20} when ± is minus. Subtract i\sqrt{215} from 5.

$$x=\frac{-\sqrt{215}i+5}{20}$$

Divide 5-i\sqrt{215} by 20.

$$x=-\frac{\sqrt{215}i}{20}+\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{215}i}{20}+\frac{1}{4} x=-\frac{\sqrt{215}i}{20}+\frac{1}{4}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$10x^{2}-2x+2-3x=-4$$

Combine -2x and -3x to get -5x.

$$10x^{2}-5x+2=-4$$

Subtract 2 from both sides.

$$10x^{2}-5x=-4-2$$

Subtract 2 from -4 to get -6.

$$10x^{2}-5x=-6$$

Divide both sides by 10.

$$\frac{10x^{2}-5x}{10}=-\frac{6}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{5}{10}\right)x=-\frac{6}{10}$$

Reduce the fraction \frac{-5}{10} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{2}x=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{3}{5}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{3}{5}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{3}{5}+\frac{1}{16}$$

Add -\frac{3}{5} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{43}{80}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{43}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{43}{80}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{215}i}{20} x-\frac{1}{4}=-\frac{\sqrt{215}i}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{215}i}{20}+\frac{1}{4} x=-\frac{\sqrt{215}i}{20}+\frac{1}{4}$$