Solve 5x^2=24x+5

$$5 x ^ { 2 } = 24 x + 5$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=5$$

Steps Using Factoring By Grouping

Subtract 24x from both sides.

$$5x^{2}-24x=5$$

Subtract 5 from both sides.

$$5x^{2}-24x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-24 ab=5\left(-5\right)=-25$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -25.

$$1,-25 5,-5$$

Calculate the sum for each pair.

$$1-25=-24 5-5=0$$

The solution is the pair that gives sum -24.

$$a=-25 b=1$$

Rewrite 5x^{2}-24x-5 as \left(5x^{2}-25x\right)+\left(x-5\right).

$$\left(5x^{2}-25x\right)+\left(x-5\right)$$

Factor out 5x in 5x^{2}-25x.

$$5x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(5x+1\right)$$

To find equation solutions, solve x-5=0 and 5x+1=0.

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

Steps Using the Quadratic Formula

Subtract 24x from both sides.

$$5x^{2}-24x=5$$

Subtract 5 from both sides.

$$5x^{2}-24x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -24 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-24\right)±\sqrt{576-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{-\left(-24\right)±\sqrt{576+100}}{2\times 5}$$

Add 576 to 100.

$$x=\frac{-\left(-24\right)±\sqrt{676}}{2\times 5}$$

Take the square root of 676.

$$x=\frac{-\left(-24\right)±26}{2\times 5}$$

The opposite of -24 is 24.

$$x=\frac{24±26}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{24±26}{10}$$

Now solve the equation x=\frac{24±26}{10} when ± is plus. Add 24 to 26.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{24±26}{10} when ± is minus. Subtract 26 from 24.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 24x from both sides.

$$5x^{2}-24x=5$$

Divide both sides by 5.

$$\frac{5x^{2}-24x}{5}=\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{24}{5}x=\frac{5}{5}$$

Divide 5 by 5.

$$x^{2}-\frac{24}{5}x=1$$

Divide -\frac{24}{5}, the coefficient of the x term, by 2 to get -\frac{12}{5}. Then add the square of -\frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{5}x+\left(-\frac{12}{5}\right)^{2}=1+\left(-\frac{12}{5}\right)^{2}$$

Square -\frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=1+\frac{144}{25}$$

Add 1 to \frac{144}{25}.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=\frac{169}{25}$$

Factor x^{2}-\frac{24}{5}x+\frac{144}{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}.

$$\left(x-\frac{12}{5}\right)^{2}=\frac{169}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{5}\right)^{2}}=\sqrt{\frac{169}{25}}$$

Simplify.

$$x-\frac{12}{5}=\frac{13}{5} x-\frac{12}{5}=-\frac{13}{5}$$

Add \frac{12}{5} to both sides of the equation.

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