Solve 5(x-3)^2+2x(x+3)=(x+2)^2+11

$$5 ( x - 3 ) ^ { 2 } + 2 x ( x + 3 ) = ( x + 2 ) ^ { 2 } + 11$$

Solve for x

$$x=3$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$5\left(x^{2}-6x+9\right)+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 5 by x^{2}-6x+9.

$$5x^{2}-30x+45+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 2x by x+3.

$$5x^{2}-30x+45+2x^{2}+6x=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-30x+45+6x=\left(x+2\right)^{2}+11$$

Combine -30x and 6x to get -24x.

$$7x^{2}-24x+45=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-24x+45=x^{2}+4x+4+11$$

Add 4 and 11 to get 15.

$$7x^{2}-24x+45=x^{2}+4x+15$$

Subtract x^{2} from both sides.

$$7x^{2}-24x+45-x^{2}=4x+15$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-24x+45=4x+15$$

Subtract 4x from both sides.

$$6x^{2}-24x+45-4x=15$$

Combine -24x and -4x to get -28x.

$$6x^{2}-28x+45=15$$

Subtract 15 from both sides.

$$6x^{2}-28x+45-15=0$$

Subtract 15 from 45 to get 30.

$$6x^{2}-28x+30=0$$

Divide both sides by 2.

$$3x^{2}-14x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-14 ab=3\times 15=45$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite 3x^{2}-14x+15 as \left(3x^{2}-9x\right)+\left(-5x+15\right).

$$\left(3x^{2}-9x\right)+\left(-5x+15\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(x-3\right)-5\left(x-3\right)$$

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

$$\left(x-3\right)\left(3x-5\right)$$

To find equation solutions, solve x-3=0 and 3x-5=0.

$$x=3 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$5\left(x^{2}-6x+9\right)+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 5 by x^{2}-6x+9.

$$5x^{2}-30x+45+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 2x by x+3.

$$5x^{2}-30x+45+2x^{2}+6x=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-30x+45+6x=\left(x+2\right)^{2}+11$$

Combine -30x and 6x to get -24x.

$$7x^{2}-24x+45=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-24x+45=x^{2}+4x+4+11$$

Add 4 and 11 to get 15.

$$7x^{2}-24x+45=x^{2}+4x+15$$

Subtract x^{2} from both sides.

$$7x^{2}-24x+45-x^{2}=4x+15$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-24x+45=4x+15$$

Subtract 4x from both sides.

$$6x^{2}-24x+45-4x=15$$

Combine -24x and -4x to get -28x.

$$6x^{2}-28x+45=15$$

Subtract 15 from both sides.

$$6x^{2}-28x+45-15=0$$

Subtract 15 from 45 to get 30.

$$6x^{2}-28x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -28 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 6\times 30}}{2\times 6}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 6\times 30}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-28\right)±\sqrt{784-24\times 30}}{2\times 6}$$

Multiply -24 times 30.

$$x=\frac{-\left(-28\right)±\sqrt{784-720}}{2\times 6}$$

Add 784 to -720.

$$x=\frac{-\left(-28\right)±\sqrt{64}}{2\times 6}$$

Take the square root of 64.

$$x=\frac{-\left(-28\right)±8}{2\times 6}$$

The opposite of -28 is 28.

$$x=\frac{28±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{28±8}{12}$$

Now solve the equation x=\frac{28±8}{12} when ± is plus. Add 28 to 8.

$$x=\frac{36}{12}$$

Divide 36 by 12.

$$x=3$$

Now solve the equation x=\frac{28±8}{12} when ± is minus. Subtract 8 from 28.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

The equation is now solved.

$$x=3 x=\frac{5}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$5\left(x^{2}-6x+9\right)+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 5 by x^{2}-6x+9.

$$5x^{2}-30x+45+2x\left(x+3\right)=\left(x+2\right)^{2}+11$$

Use the distributive property to multiply 2x by x+3.

$$5x^{2}-30x+45+2x^{2}+6x=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-30x+45+6x=\left(x+2\right)^{2}+11$$

Combine -30x and 6x to get -24x.

$$7x^{2}-24x+45=\left(x+2\right)^{2}+11$$

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

$$7x^{2}-24x+45=x^{2}+4x+4+11$$

Add 4 and 11 to get 15.

$$7x^{2}-24x+45=x^{2}+4x+15$$

Subtract x^{2} from both sides.

$$7x^{2}-24x+45-x^{2}=4x+15$$

Combine 7x^{2} and -x^{2} to get 6x^{2}.

$$6x^{2}-24x+45=4x+15$$

Subtract 4x from both sides.

$$6x^{2}-24x+45-4x=15$$

Combine -24x and -4x to get -28x.

$$6x^{2}-28x+45=15$$

Subtract 45 from both sides.

$$6x^{2}-28x=15-45$$

Subtract 45 from 15 to get -30.

$$6x^{2}-28x=-30$$

Divide both sides by 6.

$$\frac{6x^{2}-28x}{6}=-\frac{30}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{28}{6}\right)x=-\frac{30}{6}$$

Reduce the fraction \frac{-28}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{14}{3}x=-\frac{30}{6}$$

Divide -30 by 6.

$$x^{2}-\frac{14}{3}x=-5$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=-5+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=-5+\frac{49}{9}$$

Add -5 to \frac{49}{9}.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{7}{3} to both sides of the equation.

$$x=3 x=\frac{5}{3}$$