Solve left(2x-3right)^2=49

$${ \left(2x-3 \right) }^{ 2 } =49$$

Solve for x

$$x=5$$

$$x=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=49$$

Subtract 49 from both sides.

$$4x^{2}-12x+9-49=0$$

Subtract 49 from 9 to get -40.

$$4x^{2}-12x-40=0$$

Divide both sides by 4.

$$x^{2}-3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite x^{2}-3x-10 as \left(x^{2}-5x\right)+\left(2x-10\right).

$$\left(x^{2}-5x\right)+\left(2x-10\right)$$

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

$$x\left(x-5\right)+2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=49$$

Subtract 49 from both sides.

$$4x^{2}-12x+9-49=0$$

Subtract 49 from 9 to get -40.

$$4x^{2}-12x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-40\right)}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-40\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-40\right)}}{2\times 4}$$

Multiply -16 times -40.

$$x=\frac{-\left(-12\right)±\sqrt{144+640}}{2\times 4}$$

Add 144 to 640.

$$x=\frac{-\left(-12\right)±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$x=\frac{-\left(-12\right)±28}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±28}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±28}{8}$$

Now solve the equation x=\frac{12±28}{8} when ± is plus. Add 12 to 28.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{12±28}{8} when ± is minus. Subtract 28 from 12.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=5 x=-2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9=49$$

Subtract 9 from both sides.

$$4x^{2}-12x=49-9$$

Subtract 9 from 49 to get 40.

$$4x^{2}-12x=40$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=\frac{40}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{40}{4}$$

Divide -12 by 4.

$$x^{2}-3x=\frac{40}{4}$$

Divide 40 by 4.

$$x^{2}-3x=10$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=5 x=-2$$