Solve 5x^2=5-2x

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

Solve for x

$$x=\frac{\sqrt{26}-1}{5}\approx 0.819803903$$

$$x=\frac{-\sqrt{26}-1}{5}\approx -1.219803903$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

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

Add 2x to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 2.

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

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{-2±\sqrt{4+100}}{2\times 5}$$

Add 4 to 100.

$$x=\frac{-2±\sqrt{104}}{2\times 5}$$

Take the square root of 104.

$$x=\frac{-2±2\sqrt{26}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±2\sqrt{26}}{10}$$

Now solve the equation x=\frac{-2±2\sqrt{26}}{10} when ± is plus. Add -2 to 2\sqrt{26}.

$$x=\frac{2\sqrt{26}-2}{10}$$

Divide -2+2\sqrt{26} by 10.

$$x=\frac{\sqrt{26}-1}{5}$$

Now solve the equation x=\frac{-2±2\sqrt{26}}{10} when ± is minus. Subtract 2\sqrt{26} from -2.

$$x=\frac{-2\sqrt{26}-2}{10}$$

Divide -2-2\sqrt{26} by 10.

$$x=\frac{-\sqrt{26}-1}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{26}-1}{5} x=\frac{-\sqrt{26}-1}{5}$$

Steps for Completing the Square

Add 2x to both sides.

$$5x^{2}+2x=5$$

Divide both sides by 5.

$$\frac{5x^{2}+2x}{5}=\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{2}{5}x=\frac{5}{5}$$

Divide 5 by 5.

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

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=1+\frac{1}{25}$$

Add 1 to \frac{1}{25}.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{26}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{26}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{26}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{\sqrt{26}}{5} x+\frac{1}{5}=-\frac{\sqrt{26}}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{\sqrt{26}-1}{5} x=\frac{-\sqrt{26}-1}{5}$$