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

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

Solve for x

$$x = -\frac{15}{2} = -7\frac{1}{2} = -7.5$$

$$x=-1$$

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}.

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

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

$$2x^{2}+12x+18+5x=3$$

Combine 12x and 5x to get 17x.

$$2x^{2}+17x+18=3$$

Subtract 3 from both sides.

$$2x^{2}+17x+18-3=0$$

Subtract 3 from 18 to get 15.

$$2x^{2}+17x+15=0$$

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

$$a+b=17 ab=2\times 15=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 17.

$$a=2 b=15$$

Rewrite 2x^{2}+17x+15 as \left(2x^{2}+2x\right)+\left(15x+15\right).

$$\left(2x^{2}+2x\right)+\left(15x+15\right)$$

Factor out 2x in the first and 15 in the second group.

$$2x\left(x+1\right)+15\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(2x+15\right)$$

To find equation solutions, solve x+1=0 and 2x+15=0.

$$x=-1 x=-\frac{15}{2}$$

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}.

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

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

$$2x^{2}+12x+18+5x=3$$

Combine 12x and 5x to get 17x.

$$2x^{2}+17x+18=3$$

Subtract 3 from both sides.

$$2x^{2}+17x+18-3=0$$

Subtract 3 from 18 to get 15.

$$2x^{2}+17x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 17 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 2\times 15}}{2\times 2}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 2\times 15}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-17±\sqrt{289-8\times 15}}{2\times 2}$$

Multiply -8 times 15.

$$x=\frac{-17±\sqrt{289-120}}{2\times 2}$$

Add 289 to -120.

$$x=\frac{-17±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-17±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-17±13}{4}$$

Now solve the equation x=\frac{-17±13}{4} when ± is plus. Add -17 to 13.

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

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-17±13}{4} when ± is minus. Subtract 13 from -17.

$$x=-\frac{30}{4}$$

Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{15}{2}$$

The equation is now solved.

$$x=-1 x=-\frac{15}{2}$$

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}.

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

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

$$2x^{2}+12x+18+5x=3$$

Combine 12x and 5x to get 17x.

$$2x^{2}+17x+18=3$$

Subtract 18 from both sides.

$$2x^{2}+17x=3-18$$

Subtract 18 from 3 to get -15.

$$2x^{2}+17x=-15$$

Divide both sides by 2.

$$\frac{2x^{2}+17x}{2}=-\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{17}{2}x=-\frac{15}{2}$$

Divide \frac{17}{2}, the coefficient of the x term, by 2 to get \frac{17}{4}. Then add the square of \frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{2}x+\left(\frac{17}{4}\right)^{2}=-\frac{15}{2}+\left(\frac{17}{4}\right)^{2}$$

Square \frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=-\frac{15}{2}+\frac{289}{16}$$

Add -\frac{15}{2} to \frac{289}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{17}{2}x+\frac{289}{16}=\frac{169}{16}$$

Factor x^{2}+\frac{17}{2}x+\frac{289}{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}.

$$\left(x+\frac{17}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{17}{4}=\frac{13}{4} x+\frac{17}{4}=-\frac{13}{4}$$

Subtract \frac{17}{4} from both sides of the equation.

$$x=-1 x=-\frac{15}{2}$$