Solve 4(x-3)^2-28(x-3)=-49

$$4 ( x - 3 ) ^ { 2 } - 28 ( x - 3 ) = - 49$$

Solve for x

$$x = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

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}.

$$4\left(x^{2}-6x+9\right)-28\left(x-3\right)=-49$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36-28\left(x-3\right)=-49$$

Use the distributive property to multiply -28 by x-3.

$$4x^{2}-24x+36-28x+84=-49$$

Combine -24x and -28x to get -52x.

$$4x^{2}-52x+36+84=-49$$

Add 36 and 84 to get 120.

$$4x^{2}-52x+120=-49$$

Add 49 to both sides.

$$4x^{2}-52x+120+49=0$$

Add 120 and 49 to get 169.

$$4x^{2}-52x+169=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+169. To find a and b, set up a system to be solved.

$$a+b=-52 ab=4\times 169=676$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 676.

$$-1,-676 -2,-338 -4,-169 -13,-52 -26,-26$$

Calculate the sum for each pair.

$$-1-676=-677 -2-338=-340 -4-169=-173 -13-52=-65 -26-26=-52$$

The solution is the pair that gives sum -52.

$$a=-26 b=-26$$

Rewrite 4x^{2}-52x+169 as \left(4x^{2}-26x\right)+\left(-26x+169\right).

$$\left(4x^{2}-26x\right)+\left(-26x+169\right)$$

Factor out 2x in the first and -13 in the second group.

$$2x\left(2x-13\right)-13\left(2x-13\right)$$

Factor out common term 2x-13 by using distributive property.

$$\left(2x-13\right)\left(2x-13\right)$$

Rewrite as a binomial square.

$$\left(2x-13\right)^{2}$$

To find equation solution, solve 2x-13=0.

$$x=\frac{13}{2}$$

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}.

$$4\left(x^{2}-6x+9\right)-28\left(x-3\right)=-49$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36-28\left(x-3\right)=-49$$

Use the distributive property to multiply -28 by x-3.

$$4x^{2}-24x+36-28x+84=-49$$

Combine -24x and -28x to get -52x.

$$4x^{2}-52x+36+84=-49$$

Add 36 and 84 to get 120.

$$4x^{2}-52x+120=-49$$

Add 49 to both sides.

$$4x^{2}-52x+120+49=0$$

Add 120 and 49 to get 169.

$$4x^{2}-52x+169=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -52 for b, and 169 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 4\times 169}}{2\times 4}$$

Square -52.

$$x=\frac{-\left(-52\right)±\sqrt{2704-4\times 4\times 169}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-52\right)±\sqrt{2704-16\times 169}}{2\times 4}$$

Multiply -16 times 169.

$$x=\frac{-\left(-52\right)±\sqrt{2704-2704}}{2\times 4}$$

Add 2704 to -2704.

$$x=\frac{-\left(-52\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-52}{2\times 4}$$

The opposite of -52 is 52.

$$x=\frac{52}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{52}{8}$$

Reduce the fraction \frac{52}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{13}{2}$$

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}.

$$4\left(x^{2}-6x+9\right)-28\left(x-3\right)=-49$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36-28\left(x-3\right)=-49$$

Use the distributive property to multiply -28 by x-3.

$$4x^{2}-24x+36-28x+84=-49$$

Combine -24x and -28x to get -52x.

$$4x^{2}-52x+36+84=-49$$

Add 36 and 84 to get 120.

$$4x^{2}-52x+120=-49$$

Subtract 120 from both sides.

$$4x^{2}-52x=-49-120$$

Subtract 120 from -49 to get -169.

$$4x^{2}-52x=-169$$

Divide both sides by 4.

$$\frac{4x^{2}-52x}{4}=-\frac{169}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{52}{4}\right)x=-\frac{169}{4}$$

Divide -52 by 4.

$$x^{2}-13x=-\frac{169}{4}$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-\frac{169}{4}+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=\frac{-169+169}{4}$$

Add -\frac{169}{4} to \frac{169}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-13x+\frac{169}{4}=0$$

Factor x^{2}-13x+\frac{169}{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}.

$$\left(x-\frac{13}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{13}{2}=0 x-\frac{13}{2}=0$$

Add \frac{13}{2} to both sides of the equation.

$$x=\frac{13}{2} x=\frac{13}{2}$$

The equation is now solved. Solutions are the same.

$$x=\frac{13}{2}$$