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

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

Solve for x

$$x=-7$$

$$x=\frac{1}{2}=0.5$$

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}.

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

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

$$2x^{2}+8x+8=5\left(3-x\right)$$

Use the distributive property to multiply 5 by 3-x.

$$2x^{2}+8x+8=15-5x$$

Subtract 15 from both sides.

$$2x^{2}+8x+8-15=-5x$$

Subtract 15 from 8 to get -7.

$$2x^{2}+8x-7=-5x$$

Add 5x to both sides.

$$2x^{2}+8x-7+5x=0$$

Combine 8x and 5x to get 13x.

$$2x^{2}+13x-7=0$$

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

$$a+b=13 ab=2\left(-7\right)=-14$$

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 -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 13.

$$a=-1 b=14$$

Rewrite 2x^{2}+13x-7 as \left(2x^{2}-x\right)+\left(14x-7\right).

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

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

$$x\left(2x-1\right)+7\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+7\right)$$

To find equation solutions, solve 2x-1=0 and x+7=0.

$$x=\frac{1}{2} x=-7$$

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}.

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

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

$$2x^{2}+8x+8=5\left(3-x\right)$$

Use the distributive property to multiply 5 by 3-x.

$$2x^{2}+8x+8=15-5x$$

Subtract 15 from both sides.

$$2x^{2}+8x+8-15=-5x$$

Subtract 15 from 8 to get -7.

$$2x^{2}+8x-7=-5x$$

Add 5x to both sides.

$$2x^{2}+8x-7+5x=0$$

Combine 8x and 5x to get 13x.

$$2x^{2}+13x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 13 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 2\left(-7\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\left(-7\right)}}{2\times 2}$$

Multiply -8 times -7.

$$x=\frac{-13±\sqrt{169+56}}{2\times 2}$$

Add 169 to 56.

$$x=\frac{-13±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$x=\frac{-13±15}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±15}{4}$$

Now solve the equation x=\frac{-13±15}{4} when ± is plus. Add -13 to 15.

$$x=\frac{2}{4}$$

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

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-13±15}{4} when ± is minus. Subtract 15 from -13.

$$x=-\frac{28}{4}$$

Divide -28 by 4.

$$x=-7$$

The equation is now solved.

$$x=\frac{1}{2} x=-7$$

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}.

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

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

$$2x^{2}+8x+8=5\left(3-x\right)$$

Use the distributive property to multiply 5 by 3-x.

$$2x^{2}+8x+8=15-5x$$

Add 5x to both sides.

$$2x^{2}+8x+8+5x=15$$

Combine 8x and 5x to get 13x.

$$2x^{2}+13x+8=15$$

Subtract 8 from both sides.

$$2x^{2}+13x=15-8$$

Subtract 8 from 15 to get 7.

$$2x^{2}+13x=7$$

Divide both sides by 2.

$$\frac{2x^{2}+13x}{2}=\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{13}{2}x=\frac{7}{2}$$

Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=\frac{7}{2}+\left(\frac{13}{4}\right)^{2}$$

Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{7}{2}+\frac{169}{16}$$

Add \frac{7}{2} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{225}{16}$$

Factor x^{2}+\frac{13}{2}x+\frac{169}{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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{15}{4} x+\frac{13}{4}=-\frac{15}{4}$$

Subtract \frac{13}{4} from both sides of the equation.

$$x=\frac{1}{2} x=-7$$