Solve 5(x+2)^2=(7x+3)(x+2)

$$5 ( x + 2 ) ^ { 2 } = ( 7 x + 3 ) ( x + 2 )$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$5\left(x^{2}+4x+4\right)=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 5 by x^{2}+4x+4.

$$5x^{2}+20x+20=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 7x+3 by x+2 and combine like terms.

$$5x^{2}+20x+20=7x^{2}+17x+6$$

Subtract 7x^{2} from both sides.

$$5x^{2}+20x+20-7x^{2}=17x+6$$

Combine 5x^{2} and -7x^{2} to get -2x^{2}.

$$-2x^{2}+20x+20=17x+6$$

Subtract 17x from both sides.

$$-2x^{2}+20x+20-17x=6$$

Combine 20x and -17x to get 3x.

$$-2x^{2}+3x+20=6$$

Subtract 6 from both sides.

$$-2x^{2}+3x+20-6=0$$

Subtract 6 from 20 to get 14.

$$-2x^{2}+3x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=3 ab=-2\times 14=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=7 b=-4$$

Rewrite -2x^{2}+3x+14 as \left(-2x^{2}+7x\right)+\left(-4x+14\right).

$$\left(-2x^{2}+7x\right)+\left(-4x+14\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(2x-7\right)-2\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(-x-2\right)$$

To find equation solutions, solve 2x-7=0 and -x-2=0.

$$x=\frac{7}{2} x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$5\left(x^{2}+4x+4\right)=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 5 by x^{2}+4x+4.

$$5x^{2}+20x+20=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 7x+3 by x+2 and combine like terms.

$$5x^{2}+20x+20=7x^{2}+17x+6$$

Subtract 7x^{2} from both sides.

$$5x^{2}+20x+20-7x^{2}=17x+6$$

Combine 5x^{2} and -7x^{2} to get -2x^{2}.

$$-2x^{2}+20x+20=17x+6$$

Subtract 17x from both sides.

$$-2x^{2}+20x+20-17x=6$$

Combine 20x and -17x to get 3x.

$$-2x^{2}+3x+20=6$$

Subtract 6 from both sides.

$$-2x^{2}+3x+20-6=0$$

Subtract 6 from 20 to get 14.

$$-2x^{2}+3x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 3 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-2\right)\times 14}}{2\left(-2\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\times 14}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8\times 14}}{2\left(-2\right)}$$

Multiply 8 times 14.

$$x=\frac{-3±\sqrt{9+112}}{2\left(-2\right)}$$

Add 9 to 112.

$$x=\frac{-3±\sqrt{121}}{2\left(-2\right)}$$

Take the square root of 121.

$$x=\frac{-3±11}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±11}{-4}$$

Now solve the equation x=\frac{-3±11}{-4} when ± is plus. Add -3 to 11.

$$x=\frac{8}{-4}$$

Divide 8 by -4.

$$x=-2$$

Now solve the equation x=\frac{-3±11}{-4} when ± is minus. Subtract 11 from -3.

$$x=-\frac{14}{-4}$$

Reduce the fraction \frac{-14}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{2}$$

The equation is now solved.

$$x=-2 x=\frac{7}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$5\left(x^{2}+4x+4\right)=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 5 by x^{2}+4x+4.

$$5x^{2}+20x+20=\left(7x+3\right)\left(x+2\right)$$

Use the distributive property to multiply 7x+3 by x+2 and combine like terms.

$$5x^{2}+20x+20=7x^{2}+17x+6$$

Subtract 7x^{2} from both sides.

$$5x^{2}+20x+20-7x^{2}=17x+6$$

Combine 5x^{2} and -7x^{2} to get -2x^{2}.

$$-2x^{2}+20x+20=17x+6$$

Subtract 17x from both sides.

$$-2x^{2}+20x+20-17x=6$$

Combine 20x and -17x to get 3x.

$$-2x^{2}+3x+20=6$$

Subtract 20 from both sides.

$$-2x^{2}+3x=6-20$$

Subtract 20 from 6 to get -14.

$$-2x^{2}+3x=-14$$

Divide both sides by -2.

$$\frac{-2x^{2}+3x}{-2}=-\frac{14}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{3}{-2}x=-\frac{14}{-2}$$

Divide 3 by -2.

$$x^{2}-\frac{3}{2}x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x^{2}-\frac{3}{2}x=7$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=7+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=7+\frac{9}{16}$$

Add 7 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{11}{4} x-\frac{3}{4}=-\frac{11}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{7}{2} x=-2$$