Solve 10r^2-21r=-4r+6

$$10 r ^ { 2 } - 21 r = - 4 r + 6$$

Solve for r

$$r=-\frac{3}{10}=-0.3$$

$$r=2$$

Steps Using Factoring By Grouping

Add 4r to both sides.

$$10r^{2}-21r+4r=6$$

Combine -21r and 4r to get -17r.

$$10r^{2}-17r=6$$

Subtract 6 from both sides.

$$10r^{2}-17r-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10r^{2}+ar+br-6. To find a and b, set up a system to be solved.

$$a+b=-17 ab=10\left(-6\right)=-60$$

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 10r^{2}-17r-6 as \left(10r^{2}-20r\right)+\left(3r-6\right).

$$\left(10r^{2}-20r\right)+\left(3r-6\right)$$

Factor out 10r in the first and 3 in the second group.

$$10r\left(r-2\right)+3\left(r-2\right)$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(10r+3\right)$$

To find equation solutions, solve r-2=0 and 10r+3=0.

$$r=2 r=-\frac{3}{10}$$

Steps Using the Quadratic Formula

Add 4r to both sides.

$$10r^{2}-21r+4r=6$$

Combine -21r and 4r to get -17r.

$$10r^{2}-17r=6$$

Subtract 6 from both sides.

$$10r^{2}-17r-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -17 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 10\left(-6\right)}}{2\times 10}$$

Square -17.

$$r=\frac{-\left(-17\right)±\sqrt{289-4\times 10\left(-6\right)}}{2\times 10}$$

Multiply -4 times 10.

$$r=\frac{-\left(-17\right)±\sqrt{289-40\left(-6\right)}}{2\times 10}$$

Multiply -40 times -6.

$$r=\frac{-\left(-17\right)±\sqrt{289+240}}{2\times 10}$$

Add 289 to 240.

$$r=\frac{-\left(-17\right)±\sqrt{529}}{2\times 10}$$

Take the square root of 529.

$$r=\frac{-\left(-17\right)±23}{2\times 10}$$

The opposite of -17 is 17.

$$r=\frac{17±23}{2\times 10}$$

Multiply 2 times 10.

$$r=\frac{17±23}{20}$$

Now solve the equation r=\frac{17±23}{20} when ± is plus. Add 17 to 23.

$$r=\frac{40}{20}$$

Divide 40 by 20.

$$r=2$$

Now solve the equation r=\frac{17±23}{20} when ± is minus. Subtract 23 from 17.

$$r=-\frac{6}{20}$$

Reduce the fraction \frac{-6}{20} to lowest terms by extracting and canceling out 2.

$$r=-\frac{3}{10}$$

The equation is now solved.

$$r=2 r=-\frac{3}{10}$$

Steps for Completing the Square

Add 4r to both sides.

$$10r^{2}-21r+4r=6$$

Combine -21r and 4r to get -17r.

$$10r^{2}-17r=6$$

Divide both sides by 10.

$$\frac{10r^{2}-17r}{10}=\frac{6}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$r^{2}-\frac{17}{10}r=\frac{6}{10}$$

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

$$r^{2}-\frac{17}{10}r=\frac{3}{5}$$

Divide -\frac{17}{10}, the coefficient of the x term, by 2 to get -\frac{17}{20}. Then add the square of -\frac{17}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{17}{10}r+\left(-\frac{17}{20}\right)^{2}=\frac{3}{5}+\left(-\frac{17}{20}\right)^{2}$$

Square -\frac{17}{20} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{17}{10}r+\frac{289}{400}=\frac{3}{5}+\frac{289}{400}$$

Add \frac{3}{5} to \frac{289}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{17}{10}r+\frac{289}{400}=\frac{529}{400}$$

Factor r^{2}-\frac{17}{10}r+\frac{289}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{17}{20}\right)^{2}=\frac{529}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{17}{20}\right)^{2}}=\sqrt{\frac{529}{400}}$$

Simplify.

$$r-\frac{17}{20}=\frac{23}{20} r-\frac{17}{20}=-\frac{23}{20}$$

Add \frac{17}{20} to both sides of the equation.

$$r=2 r=-\frac{3}{10}$$