Solve 2x^2+28x+41=-7

$$2 x ^ { 2 } + 28 x + 41 = - 7$$

Solve for x

$$x=-12$$

$$x=-2$$

Steps Using Factoring By Grouping

Add 7 to both sides.

$$2x^{2}+28x+41+7=0$$

Add 41 and 7 to get 48.

$$2x^{2}+28x+48=0$$

Divide both sides by 2.

$$x^{2}+14x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 24=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 14.

$$a=2 b=12$$

Rewrite x^{2}+14x+24 as \left(x^{2}+2x\right)+\left(12x+24\right).

$$\left(x^{2}+2x\right)+\left(12x+24\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x+2\right)+12\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+12\right)$$

To find equation solutions, solve x+2=0 and x+12=0.

$$x=-2 x=-12$$

Steps Using the Quadratic Formula

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.

$$2x^{2}+28x+41=-7$$

Add 7 to both sides of the equation.

$$2x^{2}+28x+41-\left(-7\right)=-7-\left(-7\right)$$

Subtracting -7 from itself leaves 0.

$$2x^{2}+28x+41-\left(-7\right)=0$$

Subtract -7 from 41.

$$2x^{2}+28x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 28 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 2\times 48}}{2\times 2}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 2\times 48}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-28±\sqrt{784-8\times 48}}{2\times 2}$$

Multiply -8 times 48.

$$x=\frac{-28±\sqrt{784-384}}{2\times 2}$$

Add 784 to -384.

$$x=\frac{-28±\sqrt{400}}{2\times 2}$$

Take the square root of 400.

$$x=\frac{-28±20}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-28±20}{4}$$

Now solve the equation x=\frac{-28±20}{4} when ± is plus. Add -28 to 20.

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

Divide -8 by 4.

$$x=-2$$

Now solve the equation x=\frac{-28±20}{4} when ± is minus. Subtract 20 from -28.

$$x=-\frac{48}{4}$$

Divide -48 by 4.

$$x=-12$$

The equation is now solved.

$$x=-2 x=-12$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+28x+41=-7$$

Subtract 41 from both sides of the equation.

$$2x^{2}+28x+41-41=-7-41$$

Subtracting 41 from itself leaves 0.

$$2x^{2}+28x=-7-41$$

Subtract 41 from -7.

$$2x^{2}+28x=-48$$

Divide both sides by 2.

$$\frac{2x^{2}+28x}{2}=-\frac{48}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{28}{2}x=-\frac{48}{2}$$

Divide 28 by 2.

$$x^{2}+14x=-\frac{48}{2}$$

Divide -48 by 2.

$$x^{2}+14x=-24$$

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

$$x^{2}+14x+7^{2}=-24+7^{2}$$

Square 7.

$$x^{2}+14x+49=-24+49$$

Add -24 to 49.

$$x^{2}+14x+49=25$$

Factor x^{2}+14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+7\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+7=5 x+7=-5$$

Subtract 7 from both sides of the equation.

$$x=-2 x=-12$$