Solve 3r^2+40=-23r

$$3 r ^ { 2 } + 40 = - 23 r$$

Solve for r

$$r = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$r=-5$$

Steps Using Factoring By Grouping

Add 23r to both sides.

$$3r^{2}+40+23r=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3r^{2}+23r+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3r^{2}+ar+br+40. To find a and b, set up a system to be solved.

$$a+b=23 ab=3\times 40=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=8 b=15$$

Rewrite 3r^{2}+23r+40 as \left(3r^{2}+8r\right)+\left(15r+40\right).

$$\left(3r^{2}+8r\right)+\left(15r+40\right)$$

Factor out r in the first and 5 in the second group.

$$r\left(3r+8\right)+5\left(3r+8\right)$$

Factor out common term 3r+8 by using distributive property.

$$\left(3r+8\right)\left(r+5\right)$$

To find equation solutions, solve 3r+8=0 and r+5=0.

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

Steps Using the Quadratic Formula

Add 23r to both sides.

$$3r^{2}+40+23r=0$$

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.

$$3r^{2}+23r+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 23 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-23±\sqrt{23^{2}-4\times 3\times 40}}{2\times 3}$$

Square 23.

$$r=\frac{-23±\sqrt{529-4\times 3\times 40}}{2\times 3}$$

Multiply -4 times 3.

$$r=\frac{-23±\sqrt{529-12\times 40}}{2\times 3}$$

Multiply -12 times 40.

$$r=\frac{-23±\sqrt{529-480}}{2\times 3}$$

Add 529 to -480.

$$r=\frac{-23±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$r=\frac{-23±7}{2\times 3}$$

Multiply 2 times 3.

$$r=\frac{-23±7}{6}$$

Now solve the equation r=\frac{-23±7}{6} when ± is plus. Add -23 to 7.

$$r=-\frac{16}{6}$$

Reduce the fraction \frac{-16}{6} to lowest terms by extracting and canceling out 2.

$$r=-\frac{8}{3}$$

Now solve the equation r=\frac{-23±7}{6} when ± is minus. Subtract 7 from -23.

$$r=-\frac{30}{6}$$

Divide -30 by 6.

$$r=-5$$

The equation is now solved.

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

Steps for Completing the Square

Add 23r to both sides.

$$3r^{2}+40+23r=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$3r^{2}+23r=-40$$

Divide both sides by 3.

$$\frac{3r^{2}+23r}{3}=-\frac{40}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$r^{2}+\frac{23}{3}r=-\frac{40}{3}$$

Divide \frac{23}{3}, the coefficient of the x term, by 2 to get \frac{23}{6}. Then add the square of \frac{23}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+\frac{23}{3}r+\left(\frac{23}{6}\right)^{2}=-\frac{40}{3}+\left(\frac{23}{6}\right)^{2}$$

Square \frac{23}{6} by squaring both the numerator and the denominator of the fraction.

$$r^{2}+\frac{23}{3}r+\frac{529}{36}=-\frac{40}{3}+\frac{529}{36}$$

Add -\frac{40}{3} to \frac{529}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}+\frac{23}{3}r+\frac{529}{36}=\frac{49}{36}$$

Factor r^{2}+\frac{23}{3}r+\frac{529}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r+\frac{23}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+\frac{23}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$r+\frac{23}{6}=\frac{7}{6} r+\frac{23}{6}=-\frac{7}{6}$$

Subtract \frac{23}{6} from both sides of the equation.

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