Solve 5x^2+2=2x+7

$$5 { x }^{ 2 } +2=2x+7$$

Solve for x

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

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

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$5x^{2}+2-2x=7$$

Subtract 7 from both sides.

$$5x^{2}+2-2x-7=0$$

Subtract 7 from 2 to get -5.

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

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}.

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

Square -2.

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

Multiply -4 times 5.

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

Multiply -20 times -5.

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

Add 4 to 100.

$$x=\frac{-\left(-2\right)±\sqrt{104}}{2\times 5}$$

Take the square root of 104.

$$x=\frac{-\left(-2\right)±2\sqrt{26}}{2\times 5}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{26}}{2\times 5}$$

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.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x from both sides.

$$5x^{2}+2-2x=7$$

Subtract 2 from both sides.

$$5x^{2}-2x=7-2$$

Subtract 2 from 7 to get 5.

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

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.

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

Add \frac{1}{5} to both sides of the equation.

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