Solve 5x^2=17x+12

$$5 x ^ { 2 } = 17 x + 12$$

Solve for x

$$x=-\frac{3}{5}=-0.6$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract 17x from both sides.

$$5x^{2}-17x=12$$

Subtract 12 from both sides.

$$5x^{2}-17x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-17 ab=5\left(-12\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -17.

$$a=-20 b=3$$

Rewrite 5x^{2}-17x-12 as \left(5x^{2}-20x\right)+\left(3x-12\right).

$$\left(5x^{2}-20x\right)+\left(3x-12\right)$$

Factor out 5x in the first and 3 in the second group.

$$5x\left(x-4\right)+3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(5x+3\right)$$

To find equation solutions, solve x-4=0 and 5x+3=0.

$$x=4 x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Subtract 17x from both sides.

$$5x^{2}-17x=12$$

Subtract 12 from both sides.

$$5x^{2}-17x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -17 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-17\right)±\sqrt{289-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$x=\frac{-\left(-17\right)±\sqrt{289+240}}{2\times 5}$$

Add 289 to 240.

$$x=\frac{-\left(-17\right)±\sqrt{529}}{2\times 5}$$

Take the square root of 529.

$$x=\frac{-\left(-17\right)±23}{2\times 5}$$

The opposite of -17 is 17.

$$x=\frac{17±23}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{17±23}{10}$$

Now solve the equation x=\frac{17±23}{10} when ± is plus. Add 17 to 23.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{17±23}{10} when ± is minus. Subtract 23 from 17.

$$x=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=4 x=-\frac{3}{5}$$

Steps for Completing the Square

Subtract 17x from both sides.

$$5x^{2}-17x=12$$

Divide both sides by 5.

$$\frac{5x^{2}-17x}{5}=\frac{12}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{17}{5}x=\frac{12}{5}$$

Divide -\frac{17}{5}, the coefficient of the x term, by 2 to get -\frac{17}{10}. Then add the square of -\frac{17}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{5}x+\left(-\frac{17}{10}\right)^{2}=\frac{12}{5}+\left(-\frac{17}{10}\right)^{2}$$

Square -\frac{17}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{5}x+\frac{289}{100}=\frac{12}{5}+\frac{289}{100}$$

Add \frac{12}{5} to \frac{289}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{5}x+\frac{289}{100}=\frac{529}{100}$$

Factor x^{2}-\frac{17}{5}x+\frac{289}{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}.

$$\left(x-\frac{17}{10}\right)^{2}=\frac{529}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{10}\right)^{2}}=\sqrt{\frac{529}{100}}$$

Simplify.

$$x-\frac{17}{10}=\frac{23}{10} x-\frac{17}{10}=-\frac{23}{10}$$

Add \frac{17}{10} to both sides of the equation.

$$x=4 x=-\frac{3}{5}$$