Solve 3x^2+4(-x+5/2)^2=12

$$3 x ^ { 2 } + 4 ( - x + \frac { 5 } { 2 } ) ^ { 2 } = 12$$

Solve for x

$$x=1$$

$$x = \frac{13}{7} = 1\frac{6}{7} \approx 1.857142857$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+\frac{5}{2}\right)^{2}.

$$3x^{2}+4\left(\left(-x\right)^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Calculate -x to the power of 2 and get x^{2}.

$$3x^{2}+4\left(x^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Use the distributive property to multiply 4 by x^{2}+5\left(-x\right)+\frac{25}{4}.

$$3x^{2}+4x^{2}+20\left(-x\right)+25=12$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}+20\left(-x\right)+25=12$$

Subtract 12 from both sides.

$$7x^{2}+20\left(-x\right)+25-12=0$$

Subtract 12 from 25 to get 13.

$$7x^{2}+20\left(-x\right)+13=0$$

Multiply 20 and -1 to get -20.

$$7x^{2}-20x+13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+13. To find a and b, set up a system to be solved.

$$a+b=-20 ab=7\times 13=91$$

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 91.

$$-1,-91 -7,-13$$

Calculate the sum for each pair.

$$-1-91=-92 -7-13=-20$$

The solution is the pair that gives sum -20.

$$a=-13 b=-7$$

Rewrite 7x^{2}-20x+13 as \left(7x^{2}-13x\right)+\left(-7x+13\right).

$$\left(7x^{2}-13x\right)+\left(-7x+13\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(7x-13\right)-\left(7x-13\right)$$

Factor out common term 7x-13 by using distributive property.

$$\left(7x-13\right)\left(x-1\right)$$

To find equation solutions, solve 7x-13=0 and x-1=0.

$$x=\frac{13}{7} x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+\frac{5}{2}\right)^{2}.

$$3x^{2}+4\left(\left(-x\right)^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Calculate -x to the power of 2 and get x^{2}.

$$3x^{2}+4\left(x^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Use the distributive property to multiply 4 by x^{2}+5\left(-x\right)+\frac{25}{4}.

$$3x^{2}+4x^{2}+20\left(-x\right)+25=12$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}+20\left(-x\right)+25=12$$

Subtract 12 from both sides.

$$7x^{2}+20\left(-x\right)+25-12=0$$

Subtract 12 from 25 to get 13.

$$7x^{2}+20\left(-x\right)+13=0$$

Multiply 20 and -1 to get -20.

$$7x^{2}-20x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -20 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 7\times 13}}{2\times 7}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 7\times 13}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-20\right)±\sqrt{400-28\times 13}}{2\times 7}$$

Multiply -28 times 13.

$$x=\frac{-\left(-20\right)±\sqrt{400-364}}{2\times 7}$$

Add 400 to -364.

$$x=\frac{-\left(-20\right)±\sqrt{36}}{2\times 7}$$

Take the square root of 36.

$$x=\frac{-\left(-20\right)±6}{2\times 7}$$

The opposite of -20 is 20.

$$x=\frac{20±6}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{20±6}{14}$$

Now solve the equation x=\frac{20±6}{14} when ± is plus. Add 20 to 6.

$$x=\frac{26}{14}$$

Reduce the fraction \frac{26}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{7}$$

Now solve the equation x=\frac{20±6}{14} when ± is minus. Subtract 6 from 20.

$$x=\frac{14}{14}$$

Divide 14 by 14.

$$x=1$$

The equation is now solved.

$$x=\frac{13}{7} x=1$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+\frac{5}{2}\right)^{2}.

$$3x^{2}+4\left(\left(-x\right)^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Calculate -x to the power of 2 and get x^{2}.

$$3x^{2}+4\left(x^{2}+5\left(-x\right)+\frac{25}{4}\right)=12$$

Use the distributive property to multiply 4 by x^{2}+5\left(-x\right)+\frac{25}{4}.

$$3x^{2}+4x^{2}+20\left(-x\right)+25=12$$

Combine 3x^{2} and 4x^{2} to get 7x^{2}.

$$7x^{2}+20\left(-x\right)+25=12$$

Subtract 25 from both sides.

$$7x^{2}+20\left(-x\right)=12-25$$

Subtract 25 from 12 to get -13.

$$7x^{2}+20\left(-x\right)=-13$$

Multiply 20 and -1 to get -20.

$$7x^{2}-20x=-13$$

Divide both sides by 7.

$$\frac{7x^{2}-20x}{7}=-\frac{13}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{20}{7}x=-\frac{13}{7}$$

Divide -\frac{20}{7}, the coefficient of the x term, by 2 to get -\frac{10}{7}. Then add the square of -\frac{10}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{7}x+\left(-\frac{10}{7}\right)^{2}=-\frac{13}{7}+\left(-\frac{10}{7}\right)^{2}$$

Square -\frac{10}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{7}x+\frac{100}{49}=-\frac{13}{7}+\frac{100}{49}$$

Add -\frac{13}{7} to \frac{100}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{7}x+\frac{100}{49}=\frac{9}{49}$$

Factor x^{2}-\frac{20}{7}x+\frac{100}{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}.

$$\left(x-\frac{10}{7}\right)^{2}=\frac{9}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{7}\right)^{2}}=\sqrt{\frac{9}{49}}$$

Simplify.

$$x-\frac{10}{7}=\frac{3}{7} x-\frac{10}{7}=-\frac{3}{7}$$

Add \frac{10}{7} to both sides of the equation.

$$x=\frac{13}{7} x=1$$