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

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-2$$

Steps Using Factoring By Grouping

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

$$2x^{2}+6=-7x$$

Add 7x to both sides.

$$2x^{2}+6+7x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}+7x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 6=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 2x^{2}+7x+6 as \left(2x^{2}+3x\right)+\left(4x+6\right).

$$\left(2x^{2}+3x\right)+\left(4x+6\right)$$

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

$$x\left(2x+3\right)+2\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(x+2\right)$$

To find equation solutions, solve 2x+3=0 and x+2=0.

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

Steps Using the Quadratic Formula

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

$$2x^{2}+6=-7x$$

Add 7x to both sides.

$$2x^{2}+6+7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+7x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 6}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-7±\sqrt{49-48}}{2\times 2}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±1}{4}$$

Now solve the equation x=\frac{-7±1}{4} when ± is plus. Add -7 to 1.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-7±1}{4} when ± is minus. Subtract 1 from -7.

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

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

$$2x^{2}+6=-7x$$

Add 7x to both sides.

$$2x^{2}+6+7x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+7x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=-\frac{6}{2}$$

Divide -6 by 2.

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

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-3+\frac{49}{16}$$

Add -3 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{1}{4} x+\frac{7}{4}=-\frac{1}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

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