Solve 3x^2+17x=-3x+7

$$3 { x }^{ 2 } +17x=-3x+7$$

Solve for x

$$x=-7$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Add 3x to both sides.

$$3x^{2}+17x+3x=7$$

Combine 17x and 3x to get 20x.

$$3x^{2}+20x=7$$

Subtract 7 from both sides.

$$3x^{2}+20x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=20 ab=3\left(-7\right)=-21$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 20.

$$a=-1 b=21$$

Rewrite 3x^{2}+20x-7 as \left(3x^{2}-x\right)+\left(21x-7\right).

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

Factor out x in the first and 7 in the second group.

$$x\left(3x-1\right)+7\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(x+7\right)$$

To find equation solutions, solve 3x-1=0 and x+7=0.

$$x=\frac{1}{3} x=-7$$

Steps Using the Quadratic Formula

Add 3x to both sides.

$$3x^{2}+17x+3x=7$$

Combine 17x and 3x to get 20x.

$$3x^{2}+20x=7$$

Subtract 7 from both sides.

$$3x^{2}+20x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 20 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 3\left(-7\right)}}{2\times 3}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 3\left(-7\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-20±\sqrt{400-12\left(-7\right)}}{2\times 3}$$

Multiply -12 times -7.

$$x=\frac{-20±\sqrt{400+84}}{2\times 3}$$

Add 400 to 84.

$$x=\frac{-20±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$x=\frac{-20±22}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-20±22}{6}$$

Now solve the equation x=\frac{-20±22}{6} when ± is plus. Add -20 to 22.

$$x=\frac{2}{6}$$

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-20±22}{6} when ± is minus. Subtract 22 from -20.

$$x=-\frac{42}{6}$$

Divide -42 by 6.

$$x=-7$$

The equation is now solved.

$$x=\frac{1}{3} x=-7$$

Steps for Completing the Square

Add 3x to both sides.

$$3x^{2}+17x+3x=7$$

Combine 17x and 3x to get 20x.

$$3x^{2}+20x=7$$

Divide both sides by 3.

$$\frac{3x^{2}+20x}{3}=\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{20}{3}x=\frac{7}{3}$$

Divide \frac{20}{3}, the coefficient of the x term, by 2 to get \frac{10}{3}. Then add the square of \frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{20}{3}x+\left(\frac{10}{3}\right)^{2}=\frac{7}{3}+\left(\frac{10}{3}\right)^{2}$$

Square \frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{20}{3}x+\frac{100}{9}=\frac{7}{3}+\frac{100}{9}$$

Add \frac{7}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{20}{3}x+\frac{100}{9}=\frac{121}{9}$$

Factor x^{2}+\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{10}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{10}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x+\frac{10}{3}=\frac{11}{3} x+\frac{10}{3}=-\frac{11}{3}$$

Subtract \frac{10}{3} from both sides of the equation.

$$x=\frac{1}{3} x=-7$$