Solve -11x+19+(2x-3)(3x+4)=(x-1)^2+(x+1)

$$- 11 x + 19 + ( 2 x - 3 ) ( 3 x + 4 ) = ( x - 1 ) ^ { 2 } + ( x + 1 )$$

Solve for x

$$x = \frac{\sqrt{21} + 11}{10} \approx 1.558257569$$

$$x=\frac{11-\sqrt{21}}{10}\approx 0.641742431$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-3 by 3x+4 and combine like terms.

$$-11x+19+6x^{2}-x-12=\left(x-1\right)^{2}+x+1$$

Combine -11x and -x to get -12x.

$$-12x+19+6x^{2}-12=\left(x-1\right)^{2}+x+1$$

Subtract 12 from 19 to get 7.

$$-12x+7+6x^{2}=\left(x-1\right)^{2}+x+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-12x+7+6x^{2}=x^{2}-2x+1+x+1$$

Combine -2x and x to get -x.

$$-12x+7+6x^{2}=x^{2}-x+1+1$$

Add 1 and 1 to get 2.

$$-12x+7+6x^{2}=x^{2}-x+2$$

Subtract x^{2} from both sides.

$$-12x+7+6x^{2}-x^{2}=-x+2$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$-12x+7+5x^{2}=-x+2$$

Add x to both sides.

$$-12x+7+5x^{2}+x=2$$

Combine -12x and x to get -11x.

$$-11x+7+5x^{2}=2$$

Subtract 2 from both sides.

$$-11x+7+5x^{2}-2=0$$

Subtract 2 from 7 to get 5.

$$-11x+5+5x^{2}=0$$

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.

$$5x^{2}-11x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -11 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-11\right)±\sqrt{121-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$x=\frac{-\left(-11\right)±\sqrt{121-100}}{2\times 5}$$

Add 121 to -100.

$$x=\frac{-\left(-11\right)±\sqrt{21}}{2\times 5}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{21}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{11±\sqrt{21}}{10}$$

Now solve the equation x=\frac{11±\sqrt{21}}{10} when ± is plus. Add 11 to \sqrt{21}.

$$x=\frac{\sqrt{21}+11}{10}$$

Now solve the equation x=\frac{11±\sqrt{21}}{10} when ± is minus. Subtract \sqrt{21} from 11.

$$x=\frac{11-\sqrt{21}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+11}{10} x=\frac{11-\sqrt{21}}{10}$$

Steps for Completing the Square

Use the distributive property to multiply 2x-3 by 3x+4 and combine like terms.

$$-11x+19+6x^{2}-x-12=\left(x-1\right)^{2}+x+1$$

Combine -11x and -x to get -12x.

$$-12x+19+6x^{2}-12=\left(x-1\right)^{2}+x+1$$

Subtract 12 from 19 to get 7.

$$-12x+7+6x^{2}=\left(x-1\right)^{2}+x+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-12x+7+6x^{2}=x^{2}-2x+1+x+1$$

Combine -2x and x to get -x.

$$-12x+7+6x^{2}=x^{2}-x+1+1$$

Add 1 and 1 to get 2.

$$-12x+7+6x^{2}=x^{2}-x+2$$

Subtract x^{2} from both sides.

$$-12x+7+6x^{2}-x^{2}=-x+2$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$-12x+7+5x^{2}=-x+2$$

Add x to both sides.

$$-12x+7+5x^{2}+x=2$$

Combine -12x and x to get -11x.

$$-11x+7+5x^{2}=2$$

Subtract 7 from both sides.

$$-11x+5x^{2}=2-7$$

Subtract 7 from 2 to get -5.

$$-11x+5x^{2}=-5$$

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.

$$5x^{2}-11x=-5$$

Divide both sides by 5.

$$\frac{5x^{2}-11x}{5}=-\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{11}{5}x=-\frac{5}{5}$$

Divide -5 by 5.

$$x^{2}-\frac{11}{5}x=-1$$

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-1+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-1+\frac{121}{100}$$

Add -1 to \frac{121}{100}.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{21}{100}$$

Factor x^{2}-\frac{11}{5}x+\frac{121}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{21}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{21}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{\sqrt{21}}{10} x-\frac{11}{10}=-\frac{\sqrt{21}}{10}$$

Add \frac{11}{10} to both sides of the equation.

$$x=\frac{\sqrt{21}+11}{10} x=\frac{11-\sqrt{21}}{10}$$