Solve x^2+left(x+2right)left(11x-7right)=4

$${ x }^{ 2 } + \left( x+2 \right) \left( 11x-7 \right) = 4$$

Solve for x

$$x=-2$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x+2 by 11x-7 and combine like terms.

$$x^{2}+11x^{2}+15x-14=4$$

Combine x^{2} and 11x^{2} to get 12x^{2}.

$$12x^{2}+15x-14=4$$

Subtract 4 from both sides.

$$12x^{2}+15x-14-4=0$$

Subtract 4 from -14 to get -18.

$$12x^{2}+15x-18=0$$

Divide both sides by 3.

$$4x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=4\left(-6\right)=-24$$

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 -24.

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

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite 4x^{2}+5x-6 as \left(4x^{2}-3x\right)+\left(8x-6\right).

$$\left(4x^{2}-3x\right)+\left(8x-6\right)$$

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

$$x\left(4x-3\right)+2\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(x+2\right)$$

To find equation solutions, solve 4x-3=0 and x+2=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply x+2 by 11x-7 and combine like terms.

$$x^{2}+11x^{2}+15x-14=4$$

Combine x^{2} and 11x^{2} to get 12x^{2}.

$$12x^{2}+15x-14=4$$

Subtract 4 from both sides.

$$12x^{2}+15x-14-4=0$$

Subtract 4 from -14 to get -18.

$$12x^{2}+15x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 15 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 12\left(-18\right)}}{2\times 12}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 12\left(-18\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-15±\sqrt{225-48\left(-18\right)}}{2\times 12}$$

Multiply -48 times -18.

$$x=\frac{-15±\sqrt{225+864}}{2\times 12}$$

Add 225 to 864.

$$x=\frac{-15±\sqrt{1089}}{2\times 12}$$

Take the square root of 1089.

$$x=\frac{-15±33}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-15±33}{24}$$

Now solve the equation x=\frac{-15±33}{24} when ± is plus. Add -15 to 33.

$$x=\frac{18}{24}$$

Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.

$$x=\frac{3}{4}$$

Now solve the equation x=\frac{-15±33}{24} when ± is minus. Subtract 33 from -15.

$$x=-\frac{48}{24}$$

Divide -48 by 24.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply x+2 by 11x-7 and combine like terms.

$$x^{2}+11x^{2}+15x-14=4$$

Combine x^{2} and 11x^{2} to get 12x^{2}.

$$12x^{2}+15x-14=4$$

Add 14 to both sides.

$$12x^{2}+15x=4+14$$

Add 4 and 14 to get 18.

$$12x^{2}+15x=18$$

Divide both sides by 12.

$$\frac{12x^{2}+15x}{12}=\frac{18}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{15}{12}x=\frac{18}{12}$$

Reduce the fraction \frac{15}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{5}{4}x=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{5}{4}x=\frac{3}{2}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{3}{2}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{3}{2}+\frac{25}{64}$$

Add \frac{3}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{121}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{11}{8} x+\frac{5}{8}=-\frac{11}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

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