Solve (x-1)^2+5=(3x+2)(2x-3)

$$( x - 1 ) ^ { 2 } + 5 = ( 3 x + 2 ) ( 2 x - 3 )$$

Solve for x

$$x = \frac{\sqrt{249} + 3}{10} \approx 1.877973384$$

$$x=\frac{3-\sqrt{249}}{10}\approx -1.277973384$$

Steps Using the Quadratic Formula

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

$$x^{2}-2x+1+5=\left(3x+2\right)\left(2x-3\right)$$

Add 1 and 5 to get 6.

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

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

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

Subtract 6x^{2} from both sides.

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

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

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

Add 5x to both sides.

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

Combine -2x and 5x to get 3x.

$$-5x^{2}+3x+6=-6$$

Add 6 to both sides.

$$-5x^{2}+3x+6+6=0$$

Add 6 and 6 to get 12.

$$-5x^{2}+3x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 3 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-5\right)\times 12}}{2\left(-5\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-5\right)\times 12}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-3±\sqrt{9+20\times 12}}{2\left(-5\right)}$$

Multiply 20 times 12.

$$x=\frac{-3±\sqrt{9+240}}{2\left(-5\right)}$$

Add 9 to 240.

$$x=\frac{-3±\sqrt{249}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-3±\sqrt{249}}{-10}$$

Now solve the equation x=\frac{-3±\sqrt{249}}{-10} when ± is plus. Add -3 to \sqrt{249}.

$$x=\frac{\sqrt{249}-3}{-10}$$

Divide -3+\sqrt{249} by -10.

$$x=\frac{3-\sqrt{249}}{10}$$

Now solve the equation x=\frac{-3±\sqrt{249}}{-10} when ± is minus. Subtract \sqrt{249} from -3.

$$x=\frac{-\sqrt{249}-3}{-10}$$

Divide -3-\sqrt{249} by -10.

$$x=\frac{\sqrt{249}+3}{10}$$

The equation is now solved.

$$x=\frac{3-\sqrt{249}}{10} x=\frac{\sqrt{249}+3}{10}$$

Steps for Completing the Square

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

$$x^{2}-2x+1+5=\left(3x+2\right)\left(2x-3\right)$$

Add 1 and 5 to get 6.

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

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

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

Subtract 6x^{2} from both sides.

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

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

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

Add 5x to both sides.

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

Combine -2x and 5x to get 3x.

$$-5x^{2}+3x+6=-6$$

Subtract 6 from both sides.

$$-5x^{2}+3x=-6-6$$

Subtract 6 from -6 to get -12.

$$-5x^{2}+3x=-12$$

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 3 by -5.

$$x^{2}-\frac{3}{5}x=-\frac{12}{-5}$$

Divide -12 by -5.

$$x^{2}-\frac{3}{5}x=\frac{12}{5}$$

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\frac{12}{5}+\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{12}{5}+\frac{9}{100}$$

Add \frac{12}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{249}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{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}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{249}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{249}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{\sqrt{249}}{10} x-\frac{3}{10}=-\frac{\sqrt{249}}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=\frac{\sqrt{249}+3}{10} x=\frac{3-\sqrt{249}}{10}$$