Solve 5x^2-251x+3020=0

$$5 { x }^{ 2 } -251x+3020=0$$

Solve for x

$$x=20$$

$$x = \frac{151}{5} = 30\frac{1}{5} = 30.2$$

Steps Using the Quadratic Formula

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.

$$5x^{2}-251x+3020=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -251 for b, and 3020 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-251\right)±\sqrt{\left(-251\right)^{2}-4\times 5\times 3020}}{2\times 5}$$

Square -251.

$$x=\frac{-\left(-251\right)±\sqrt{63001-4\times 5\times 3020}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-251\right)±\sqrt{63001-20\times 3020}}{2\times 5}$$

Multiply -20 times 3020.

$$x=\frac{-\left(-251\right)±\sqrt{63001-60400}}{2\times 5}$$

Add 63001 to -60400.

$$x=\frac{-\left(-251\right)±\sqrt{2601}}{2\times 5}$$

Take the square root of 2601.

$$x=\frac{-\left(-251\right)±51}{2\times 5}$$

The opposite of -251 is 251.

$$x=\frac{251±51}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{251±51}{10}$$

Now solve the equation x=\frac{251±51}{10} when ± is plus. Add 251 to 51.

$$x=\frac{302}{10}$$

Reduce the fraction \frac{302}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{151}{5}$$

Now solve the equation x=\frac{251±51}{10} when ± is minus. Subtract 51 from 251.

$$x=\frac{200}{10}$$

Divide 200 by 10.

$$x=20$$

The equation is now solved.

$$x=\frac{151}{5} x=20$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$5x^{2}-251x+3020=0$$

Subtract 3020 from both sides of the equation.

$$5x^{2}-251x+3020-3020=-3020$$

Subtracting 3020 from itself leaves 0.

$$5x^{2}-251x=-3020$$

Divide both sides by 5.

$$\frac{5x^{2}-251x}{5}=-\frac{3020}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{251}{5}x=-\frac{3020}{5}$$

Divide -3020 by 5.

$$x^{2}-\frac{251}{5}x=-604$$

Divide -\frac{251}{5}, the coefficient of the x term, by 2 to get -\frac{251}{10}. Then add the square of -\frac{251}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{251}{5}x+\left(-\frac{251}{10}\right)^{2}=-604+\left(-\frac{251}{10}\right)^{2}$$

Square -\frac{251}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{251}{5}x+\frac{63001}{100}=-604+\frac{63001}{100}$$

Add -604 to \frac{63001}{100}.

$$x^{2}-\frac{251}{5}x+\frac{63001}{100}=\frac{2601}{100}$$

Factor x^{2}-\frac{251}{5}x+\frac{63001}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{251}{10}\right)^{2}=\frac{2601}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{251}{10}\right)^{2}}=\sqrt{\frac{2601}{100}}$$

Simplify.

$$x-\frac{251}{10}=\frac{51}{10} x-\frac{251}{10}=-\frac{51}{10}$$

Add \frac{251}{10} to both sides of the equation.

$$x=\frac{151}{5} x=20$$