Solve 3r^2-5(r+1)=7r+58

$$3 r ^ { 2 } - 5 ( r + 1 ) = 7 r + 58$$

Solve for r

$$r=-3$$

$$r=7$$

Steps Using Factoring By Grouping

Use the distributive property to multiply -5 by r+1.

$$3r^{2}-5r-5=7r+58$$

Subtract 7r from both sides.

$$3r^{2}-5r-5-7r=58$$

Combine -5r and -7r to get -12r.

$$3r^{2}-12r-5=58$$

Subtract 58 from both sides.

$$3r^{2}-12r-5-58=0$$

Subtract 58 from -5 to get -63.

$$3r^{2}-12r-63=0$$

Divide both sides by 3.

$$r^{2}-4r-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-21. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-21\right)=-21$$

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 -21.

$$1,-21 3,-7$$

Calculate the sum for each pair.

$$1-21=-20 3-7=-4$$

The solution is the pair that gives sum -4.

$$a=-7 b=3$$

Rewrite r^{2}-4r-21 as \left(r^{2}-7r\right)+\left(3r-21\right).

$$\left(r^{2}-7r\right)+\left(3r-21\right)$$

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

$$r\left(r-7\right)+3\left(r-7\right)$$

Factor out common term r-7 by using distributive property.

$$\left(r-7\right)\left(r+3\right)$$

To find equation solutions, solve r-7=0 and r+3=0.

$$r=7 r=-3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -5 by r+1.

$$3r^{2}-5r-5=7r+58$$

Subtract 7r from both sides.

$$3r^{2}-5r-5-7r=58$$

Combine -5r and -7r to get -12r.

$$3r^{2}-12r-5=58$$

Subtract 58 from both sides.

$$3r^{2}-12r-5-58=0$$

Subtract 58 from -5 to get -63.

$$3r^{2}-12r-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -12 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\left(-63\right)}}{2\times 3}$$

Square -12.

$$r=\frac{-\left(-12\right)±\sqrt{144-4\times 3\left(-63\right)}}{2\times 3}$$

Multiply -4 times 3.

$$r=\frac{-\left(-12\right)±\sqrt{144-12\left(-63\right)}}{2\times 3}$$

Multiply -12 times -63.

$$r=\frac{-\left(-12\right)±\sqrt{144+756}}{2\times 3}$$

Add 144 to 756.

$$r=\frac{-\left(-12\right)±\sqrt{900}}{2\times 3}$$

Take the square root of 900.

$$r=\frac{-\left(-12\right)±30}{2\times 3}$$

The opposite of -12 is 12.

$$r=\frac{12±30}{2\times 3}$$

Multiply 2 times 3.

$$r=\frac{12±30}{6}$$

Now solve the equation r=\frac{12±30}{6} when ± is plus. Add 12 to 30.

$$r=\frac{42}{6}$$

Divide 42 by 6.

$$r=7$$

Now solve the equation r=\frac{12±30}{6} when ± is minus. Subtract 30 from 12.

$$r=-\frac{18}{6}$$

Divide -18 by 6.

$$r=-3$$

The equation is now solved.

$$r=7 r=-3$$

Steps for Completing the Square

Use the distributive property to multiply -5 by r+1.

$$3r^{2}-5r-5=7r+58$$

Subtract 7r from both sides.

$$3r^{2}-5r-5-7r=58$$

Combine -5r and -7r to get -12r.

$$3r^{2}-12r-5=58$$

Add 5 to both sides.

$$3r^{2}-12r=58+5$$

Add 58 and 5 to get 63.

$$3r^{2}-12r=63$$

Divide both sides by 3.

$$\frac{3r^{2}-12r}{3}=\frac{63}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$r^{2}+\left(-\frac{12}{3}\right)r=\frac{63}{3}$$

Divide -12 by 3.

$$r^{2}-4r=\frac{63}{3}$$

Divide 63 by 3.

$$r^{2}-4r=21$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-4r+\left(-2\right)^{2}=21+\left(-2\right)^{2}$$

Square -2.

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

Add 21 to 4.

$$r^{2}-4r+4=25$$

Factor r^{2}-4r+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-2\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-2\right)^{2}}=\sqrt{25}$$

Simplify.

$$r-2=5 r-2=-5$$

Add 2 to both sides of the equation.

$$r=7 r=-3$$