Solve 4x^2+20x+25=49

$$4 x ^ { 2 } + 20 x + 25 = 49$$

Solve for x

$$x=-6$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 49 from both sides.

$$4x^{2}+20x+25-49=0$$

Subtract 49 from 25 to get -24.

$$4x^{2}+20x-24=0$$

Divide both sides by 4.

$$x^{2}+5x-6=0$$

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

$$a+b=5 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite x^{2}+5x-6 as \left(x^{2}-x\right)+\left(6x-6\right).

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

Factor out x in the first and 6 in the second group.

$$x\left(x-1\right)+6\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

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.

$$4x^{2}+20x+25=49$$

Subtract 49 from both sides of the equation.

$$4x^{2}+20x+25-49=49-49$$

Subtracting 49 from itself leaves 0.

$$4x^{2}+20x+25-49=0$$

Subtract 49 from 25.

$$4x^{2}+20x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 20 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 4\left(-24\right)}}{2\times 4}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 4\left(-24\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-20±\sqrt{400-16\left(-24\right)}}{2\times 4}$$

Multiply -16 times -24.

$$x=\frac{-20±\sqrt{400+384}}{2\times 4}$$

Add 400 to 384.

$$x=\frac{-20±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$x=\frac{-20±28}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-20±28}{8}$$

Now solve the equation x=\frac{-20±28}{8} when ± is plus. Add -20 to 28.

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

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{-20±28}{8} when ± is minus. Subtract 28 from -20.

$$x=-\frac{48}{8}$$

Divide -48 by 8.

$$x=-6$$

The equation is now solved.

$$x=1 x=-6$$

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.

$$4x^{2}+20x+25=49$$

Subtract 25 from both sides of the equation.

$$4x^{2}+20x+25-25=49-25$$

Subtracting 25 from itself leaves 0.

$$4x^{2}+20x=49-25$$

Subtract 25 from 49.

$$4x^{2}+20x=24$$

Divide both sides by 4.

$$\frac{4x^{2}+20x}{4}=\frac{24}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{20}{4}x=\frac{24}{4}$$

Divide 20 by 4.

$$x^{2}+5x=\frac{24}{4}$$

Divide 24 by 4.

$$x^{2}+5x=6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=1 x=-6$$