Solve x*(5x-7)=24

$$x \cdot ( 5 x - 7 ) = 24$$

Solve for x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 5x-7.

$$5x^{2}-7x=24$$

Subtract 24 from both sides.

$$5x^{2}-7x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\left(-24\right)}}{2\times 5}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 5\left(-24\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-7\right)±\sqrt{49-20\left(-24\right)}}{2\times 5}$$

Multiply -20 times -24.

$$x=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 5}$$

Add 49 to 480.

$$x=\frac{-\left(-7\right)±\sqrt{529}}{2\times 5}$$

Take the square root of 529.

$$x=\frac{-\left(-7\right)±23}{2\times 5}$$

The opposite of -7 is 7.

$$x=\frac{7±23}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{7±23}{10}$$

Now solve the equation x=\frac{7±23}{10} when ± is plus. Add 7 to 23.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{7±23}{10} when ± is minus. Subtract 23 from 7.

$$x=-\frac{16}{10}$$

Reduce the fraction \frac{-16}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{5}$$

The equation is now solved.

$$x=3 x=-\frac{8}{5}$$

Steps for Completing the Square

Use the distributive property to multiply x by 5x-7.

$$5x^{2}-7x=24$$

Divide both sides by 5.

$$\frac{5x^{2}-7x}{5}=\frac{24}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{7}{5}x=\frac{24}{5}$$

Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{5}x+\left(-\frac{7}{10}\right)^{2}=\frac{24}{5}+\left(-\frac{7}{10}\right)^{2}$$

Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{24}{5}+\frac{49}{100}$$

Add \frac{24}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{529}{100}$$

Factor x^{2}-\frac{7}{5}x+\frac{49}{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}.

$$\left(x-\frac{7}{10}\right)^{2}=\frac{529}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{10}\right)^{2}}=\sqrt{\frac{529}{100}}$$

Simplify.

$$x-\frac{7}{10}=\frac{23}{10} x-\frac{7}{10}=-\frac{23}{10}$$

Add \frac{7}{10} to both sides of the equation.

$$x=3 x=-\frac{8}{5}$$