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

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

Solve for x

$$x=-3$$

$$x=0$$

Steps Using Factoring

Subtract x from both sides.

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

Combine 7x and -x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$2x^{2}+6x=0$$

Factor out x.

$$x\left(2x+6\right)=0$$

To find equation solutions, solve x=0 and 2x+6=0.

$$x=0 x=-3$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine 7x and -x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$2x^{2}+6x=0$$

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

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 2}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±6}{4}$$

Now solve the equation x=\frac{-6±6}{4} when ± is plus. Add -6 to 6.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-6±6}{4} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=0 x=-3$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine 7x and -x to get 6x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

$$2x^{2}+6x=0$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{0}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{0}{2}$$

Divide 0 by 2.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=0 x=-3$$